please help asap!

3. A double replacement reaction occurs between two solutions of lead (II) nitrate and potassium bromide. Write a
balanced equation for this reaction-identifying the product that will precipitate, and the product that will remain in
solution.
a) Write the balanced equation for this double replacement reaction.
b) If this reaction starts with 32.5 g lead (II) nitrate and 38.75 g potassium bromide, how many grams of the
precipitate will be produced? Remember to use the limiting reactant to calculate the amount of precipitate
formed.
c) How many grams of the excess reactant will remain?

The oxidation number of the unknown chlorine in the compound is + 5

The oxidation number of an element in a compound is determined by a set of rules based on its position in the periodic table, as well as the charges of other atoms in the compound

We know that the oxidation number of the chlorine which we want to obtain would be designated as x and the total of the oxidation numbers of the elements in the compound is zero.

Thus we have that;

1 + x + 3(-2) = 0

1 + x - 6 = 0

-5 + x = 0

x = 5

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A silver bar with the mass of the 300 grams is heated from the 30 °C to 55 °C. The amount heat does the silver bar absorb in the joules is 1762.5 J.

The mass of the silver bar = 300 g

The initial temperature = 30 °C

The final temperature = 55 °C

The heat energy is expressed as :

Q = mc ΔT

Where,

The m is mass of the silver bar = 300 g

The c is the specific heat capacity = 0.235 J/g °C

The ΔT is the change in the temperature = final temperature - initial temperature

The ΔT is the change in the temperature = 55 °C - 30 °C

The ΔT is the change in the temperature = 25 °C

The heat energy, Q = 300 × 0.235 × 25

The heat energy, Q = 1762.5 J

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The equation that shows that the total mass during a chemical reaction stays the same is 2) H₂ + O₂ → H₂O.

This is an example of a balanced chemical equation where the number of atoms of each element on both the reactant and product side is equal. In other words, the total number of atoms of each element is conserved, and therefore the total mass is conserved. In the other reactions, either the number of atoms on the product side is different from the reactant side or there is no reaction at all.

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Answer:

Explanation:

[tex]\frac{918.7 g}{1} *\frac{1}{216.59m } = 4.241 mol[/tex] To start off the mol of HgO must be found.

After that the molar ratio between HgO and O must be found but in this case its 1:1

[tex]4.241 mol HgO*\frac{1 molO}{1molHgO} = 4.241 mol O[/tex] the mols of HgO is put on the bottom to cancel out  with the other one leaving just mols of oxygen. Finally to find g of oxygen it must be multiplied by its molar mass.

[tex]\frac{4.241 molO}{1} * \frac{15.999 g}{mol} = 67.85 g[/tex] Oxygen

Answer:

Explanation:

In the given reaction, carbon has a greater oxidation number in carbon dioxide (CO₂) than in carbon monoxide (CO). In CO₂, the oxidation number of carbon is +4, while in CO it is +2.

Explanation:

Table H lists vapor pressure data for pure water at various temperatures. We can use this data to estimate the vapor pressure of the water in the given system at its final temperature of 80.0°C.

First, we need to calculate the heat absorbed by the water sample during the heating process. We can use the specific heat capacity of water to do this:

q = m * c * ΔT

where q is the heat absorbed, m is the mass of water (100 g), c is the specific heat capacity of water, and ΔT is the temperature change (80°C - 30°C = 50°C).

Plugging in the values, we get:

q = 100 g * 4.18 J/(g*C) * 50 C

q = 20900 J

This tells us that 20,900 joules of energy were absorbed by the water sample during heating.

Next, we need to consider the saturated solution of KCIO3 in the water sample. At 80.0°C, the water is already close to boiling, so it is likely that the vapor pressure of the water in the system is close to the vapor pressure of pure water at this temperature. From Table H, we can see that the vapor pressure of pure water is approximately 356 mmHg at 80.0°C.

Therefore, the vapor pressure of the water in the given system at its final temperature of 80.0°C is approximately 356 mmHg.

The rate constant, k = 5.27 E-2 s⁻¹, determines the rate law for the reaction P → E + Z.

The rate of the reaction P → E + Z can be expressed as:

Rate = - d[P]/dt = d[E]/dt = d[Z]/dt

where d[P], d[E], and d[Z] = changes in the concentrations of P, E, and Z, respectively, over a small time interval dt.

Use the experimental data to determine the rate constant and the order of the reaction.

Calculate the initial rate of the reaction in each trial by dividing the change in concentration of P by the time interval:

rate1 = (d[P]/dt)1 = (0.30 M - 0 M)/(20 s) = 0.015 M/s

rate2 = (d[P]/dt)2 = (0.60 M - 0.30 M)/(20 s) = 0.015 M/s

rate3 = (d[P]/dt)3 = (0.90 M - 0.60 M)/(20 s) = 0.015 M/s

The initial rates are the same in all three trials, which suggests that the reaction is first-order with respect to P.

Now using any of the three trials to determine the value of the rate constant k, trial 1:

Rate1 = k[P]1

k = Rate1/[P]1 = (1.58 E-2 M/s)/(0.30 M) = 5.27 E-2 s⁻¹

Therefore, the rate law for the reaction P → E + Z is:

Rate = k[P]

where k = 5.27 E-2 s⁻¹ is the rate constant.

Use the rate law to calculate the expected rates of the reaction at different concentrations of P. For example:

Rate2 = k[P]2 = (5.27 E-2 s⁻¹)(0.60 M) = 3.16 E-2 M/s

Rate3 = k[P]3 = (5.27 E-2 s⁻¹)(0.90 M) = 4.74 E-2 M/s

These expected rates are close to the experimental rates, which suggests that the rate law is a good approximation for the reaction under these conditions.

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The nucleus completing the following equation is option C: ₂₄⁵⁰Cr.

This reaction is a type of radioactive nuclei decay.

Radioactive decay is the process by which unstable atomic nuclei undergo spontaneous transformations in order to achieve a more stable state. This is accomplished by the emission of particles and/or electromagnetic radiation from the nucleus. The decay may occur by several mechanisms, including alpha decay, beta decay, gamma decay, and electron capture.

In alpha decay, the nucleus emits an alpha particle, which consists of two protons and two neutrons, resulting in a daughter nucleus that has two fewer protons and two fewer neutrons than the original nucleus.

In beta decay, a neutron in the nucleus is converted into a proton and an electron, and the electron is then emitted from the nucleus as a beta particle. This results in the daughter nucleus having one more proton and one fewer neutron than the original nucleus.

In gamma decay, the nucleus emits a gamma ray, which is a high-energy electromagnetic radiation, without changing the number of protons or neutrons in the nucleus.

In electron capture, an electron from the inner shell of the atom is captured by the nucleus, and a proton in the nucleus is converted into a neutron. This results in the daughter nucleus having one fewer proton and one more neutron than the original nucleus.

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First, we need to write out the balanced chemical equation for the reaction: 4 KO2 + 2 CO2 → 2 K2CO3 + 3 O2

From the equation, we can see that 4 moles of KO2 react with 2 moles of CO2 to produce 3 moles of O2. Therefore, we need to convert the given masses of KO2 and CO2 into moles:

moles of KO2 = 2.61 g / molar mass of KO2 = 2.61 g / 71.10 g/mol = 0.0367 mol
moles of CO2 = 4.46 g / molar mass of CO2 = 4.46 g / 44.01 g/mol = 0.1013 mol

Next, we need to determine the limiting reagent (the reactant that will be completely consumed in the reaction) by comparing the mole ratios of KO2 and CO2 in the balanced equation. The ratio of moles of KO2 to moles of CO2 is:
0.0367 mol KO2 / 4 mol KO2 per 2 mol CO2 = 0.0184 mol CO2

Since this ratio is less than the actual number of moles of CO2 we have (0.1013 mol), CO2 is in excess and KO2 is the limiting reagent.

Using the mole ratio from the balanced equation, we can calculate the number of moles of O2 produced:

moles of O2 = 3 mol O2 per 4 mol KO2 × 0.0367 mol KO2 = 0.0275 mol O2

Finally, we can convert the moles of O2 to grams:

mass of O2 = moles of O2 × molar mass of O2 = 0.0275 mol × 32.00 g/mol = 0.88 g
Therefore, 2.61 g of KO2 and 4.46 g of CO2 would produce 0.88 g of O2.

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A sunscreen made with Benzophenone-5 ([tex]C_1_4H_1_1NaO_6S[/tex]) would be expected to be more waterproof than benzophenone-3 ([tex]C_1_4H_1_2O_3[/tex]).

This is due to the presence of a sodium salt group (Na) and a sulfonic acid group ([tex]SO_3H[/tex] ) in benzophenone-5, which makes it more polar than benzophenone-3. Polar molecules interact more strongly with water molecules and are less likely to dissolve in nonpolar solvents such as oils.

Because sunscreen is designed to be water-resistant, the more polar benzophenone-5 should have stronger interactions with water and give more water resistance than benzophenone-3.

Moreover, the sulfonic acid group in benzophenone-5 may allow it to make stronger hydrogen bonds with water, increasing its water resistance even further.

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5.221 grams of Al are required to react with 352 mL of 1.65 M HCl solution.

Molarity (M) is defined as the moles of solute per liter of the solution.

Balanced chemical equation is : 2Al + 6HCl → 2AlCl₃ + 3H₂

From the equation, we can see that 2 moles of Al react with 6 moles of HCl to produce 2 moles of AlCl₃ and 3 moles of H₂.

As moles of HCl = Molarity × Volume

moles of HCl = 1.65 mol/L × 0.352 L

moles of HCl = 0.58128 mol

and moles of Al = (2/6) × moles of HCl

moles of Al = (1/3) × 0.58128 mol

moles of Al = 0.19376 mol

mass of Al = moles of Al × molar mass of Al

mass of Al = 0.19376 mol × 26.98 g/mol

mass of Al = 5.221 g

So, 5.221 grams of Al are required to react with 352 mL of 1.65 M HCl solution.

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Answer: 0.24 Moles. WARNING: "mm" in the original question seems to be a typo and was assumed to be "M" (moles per liter) for concentration.

Explanation:

To calculate the number of moles of solute needed to prepare a solution, we can use the formula:

moles of solute = concentration (in moles per liter) * volume of solution (in liters)

Given:

Concentration of NaCl solution = 0.8 M (moles per liter)

Volume of solution to be prepared = 300 mL = 300/1000 L (converted to liters)

Plugging in the given values:

Concentration = 0.8 M

Volume of solution = 300/1000 L

moles of NaCl = 0.8 M * 300/1000 L

Calculating:

moles of NaCl = 0.24 moles

So, 0.24 moles of NaCl are needed to prepare 300 mL of a 0.8 M NaCl solution.

Answer:

0.0104 moles of gas in the flask.

Explanation:

To calculate the number of moles of gas in the flask, you can use the ideal gas law equation: PV = nRT. Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant and T is temperature.

First, you need to convert the pressure from mmHg to atm and the temperature from Celsius to Kelvin. The pressure in atm is 760.0 mmHg / 760 mmHg/atm = 1 atm. The temperature in Kelvin is 17.00°C + 273.15 = 290.15 K.

Next, you need to convert the volume from mL to L. The volume in L is 250.0 mL / 1000 mL/L = 0.2500 L.

Now you can plug all the values into the ideal gas law equation and solve for n: (1 atm)(0.2500 L) = n(0.08206 L·atm/mol·K)(290.15 K). Solving for n gives n = 0.0104 mol.

So there are approximately 0.0104 moles of gas in the flask.

The pressure of a gas that occupies 50.5L at 69.0°C is 6.95 atm.

The pressure of an ideal gas can be calculated using Avogadro's equation as follows;

PV = nRT

Where;

According to this question, 12.5 mol of an ideal gas occupies 50.5 L at 69.00°C. The pressure can be calculated as follows:

P × 50.5 = 12.5 × 0.0821 × 342

50.5P = 350.9775

P = 6.95 atm

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We can use the combined gas law to determine the volume of the balloon at a higher altitude. The combined gas law relates the pressure, volume, and temperature of a gas:

(P1 x V1) / T1 = (P2 x V2) / T2

where P1, V1, and T1 are the pressure, volume, and temperature of the gas at the initial state, and P2, V2, and T2 are the pressure, volume, and temperature of the gas at the final state.

We are given the initial pressure (P1 = 761 mmHg), volume (V1 = 56.0 L), and temperature (T1 = 23.1 °C = 296.25 K) of the gas, and the final pressure (P2 = 0.0772 atm), and temperature (T2 = -6.97 °C = 266.18 K) of the gas. We can solve for V2, the final volume of the gas:

(P1 x V1) / T1 = (P2 x V2) / T2

V2 = (P1 x V1 x T2) / (P2 x T1)

V2 = (761 mmHg x 56.0 L x 266.18 K) / (0.0772 atm x 296.25 K)

V2 = 2,040 L (rounded to three significant figures)

Therefore, the volume of the weather balloon at the higher altitude is approximately 2,040 L.

Answer:

OB. formation of new elements.

Nuclear reactions involve changes in the nucleus of an atom, such as the splitting of a nucleus or the combining of two nuclei. These reactions can result in the formation of new elements, as the number of protons in the nucleus determines the element. In contrast, chemical reactions involve the rearrangement of electrons between atoms to form new compounds, but do not involve changes to the nucleus.

A 6.50-g sample of copper metal at 25.0 °C is heated by the addition of 145 J of energy. The final temperature of the copper is 83.7 °C. The specific heat capacity of copper is 0.38 J/g-K.

The correct answer choice is "83.7"

To solve this problem, we can use the equation:

q = mcΔT

where q is the amount of energy absorbed by the copper, m is the mass of the copper, c is the specific heat capacity of copper, and ΔT is the change in temperature of the copper.

Rearranging this equation to solve for ΔT, we get:

ΔT = q / (mc)

Substituting the given values, we get:

ΔT = 145 J / (6.50 g x 0.38 J/g-K)

ΔT = 58.7 K

Therefore, the final temperature of the copper is:

25.0 °C + 58.7 °C = 83.7 °C

So the correct option is 83.7.

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The molar equilibrium concentration of Cu² ⁺ is [tex]3.3 x 10^-26[/tex]M when 0.100 mol NH₃ is added to 1.00 L of 0.800 M with Kf = [tex]2.1 x 10^13[/tex].

The issue includes computing the molar balance convergence of Cu² ⁺ in an answer that has 0.100 mol NH3 added to 1.00 L of 0.800 M [Cu(NH₃)₄]²⁺ . To tackle the issue, we can involve the balance consistent articulation for the arrangement of[Cu(NH₃)₄]²⁺:

Kf =[Cu(NH₃)₄]²⁺/(Cu² ⁺)(NH₃)₄

Since the volume of the arrangement doesn't change when NH₃ is added, the underlying grouping of [Cu(NH₃)₄]²⁺ is as yet 0.800 M, while the underlying convergences of Cu² ⁺ and NH₃ are zero. Let x be the molar centralization of Cu² ⁺ at balance, and (0.100 - 4x) be the molar grouping of NH₃ at harmony. Then, we can compose the harmony consistent articulation as:

[tex]2.1 x 10^13 = (x)(0.800 - x)^4/(0.100 - 4x)^4[/tex]

Settling for x gives the molar balance convergence of Cu² ⁺ as 3.3 x [tex]10^-26[/tex] M. This tiny focus demonstrates that the majority of the copper in the arrangement is as [Cu(NH₃)₄]²⁺ and that the expansion of NH₃ has moved the harmony towards the development of this complex.

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The molarity of the solution that has 2.0 moles of solute in 3.0 L of solution is 0.67 mol/L

Molarity is  described as a measure of the concentration of a chemical species, in particular of a solute in a solution, in terms of amount of substance per unit volume of solution.

Molarity = moles of solute / liters of solution

we then substitute the given values, and have

Molarity = 2.0 moles / 3.0 L

Molarity = 0.67 mol/L

Molarity is  very important because the ration used to express the concentration of  any  solution.

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Answer:

true

Explanation:

According to the law of conservation of mass, the mass of the reactant must be equal to the mass of the product, hence the reaction is wrong

The law of conservation of mass states that mass within a closed system remains the same over time.

It states that the mass in an isolated system can neither be created nor be destroyed but can be transformed from one form to another.

Thus,  the mass of the reactants must be equal to the mass of the products for a low energy thermodynamic process.

From the information given, we have the reaction written as;

HCI + NaOH → NaCl

The mass of the reactant Hydrogen(H) is not found on the product

The mass of the reactant(Oxygen) is also not found

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The mole fractions of each species at equilibrium are 0.25 for A and D, 0.5 for B, and 0.225 for C.

we can use the principles of chemical equilibrium and the mole fraction formula.

First, we need to write the balanced chemical equation for the reaction involving A, B, C, and D. Let's assume that the reaction is:

A + 2B <=> C + D

where A, B, C, and D are the chemical species, and the coefficients indicate their stoichiometric ratios.

Next, we need to write the expression for the equilibrium constant, Kc, for this reaction:

Kc = [C][D] / [A][B]²

where [X] denotes the molar concentration of species X at equilibrium.

Since we know the initial moles of A, B, and D, we can calculate their total moles in the mixture:

Total moles = 1 mol A + 2 mol B + 1 mol D = 4 mol

We also know that the final mixture contains 0.9 mol of C. Therefore, the molar concentration of C at equilibrium is:

[C] = 0.9 mol / 4 L = 0.225 M

Since we have only one unknown, we can use the equilibrium constant expression to calculate the molar concentration of D:

Kc = [C][D] / [A][B]²

0.9 = (0.225)(D) / (1)(2²)

D = 1.8

Therefore, the molar concentration of D at equilibrium is 1.8 M.

Using the law of conservation of mass, we can also calculate the molar concentration of A and B at equilibrium:

[A] = 1 mol / 4 L = 0.25 M

[B] = 2 mol / 4 L = 0.5 M

Mole fraction of X = moles of X / total moles

Mole fraction of A = 1 mol / 4 mol = 0.25

Mole fraction of B = 2 mol / 4 mol = 0.5

Mole fraction of C = 0.9 mol / 4 mol = 0.225

Mole fraction of D = 1 mol / 4 mol = 0.25

Therefore, the mole fractions of each species at equilibrium are 0.25 for A and D, 0.5 for B, and 0.225 for C.

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If a skimmer and a gull eat the same fish and the skimmer is more successful at catching the fish over time, it is likely that the skimmer population would increase, while the gull population may decrease.

The skimmer's success in catching the fish would give it an advantage in obtaining the necessary nutrients for survival and reproduction. As a result, the skimmer population would likely grow over time as more individuals are able to survive and reproduce due to the abundance of food.

On the other hand, the gull population may decrease due to the competition with the skimmer for the same food source. If the skimmer population grows significantly, it may lead to a reduction in the availability of fish for the gulls to feed on. Over time, this could result in a decline in the gull population due to reduced food availability.

However, it is important to note that the impact on the bird populations may depend on various factors such as the size of the populations, availability of other food sources, and environmental factors. Therefore, the outcome of this scenario cannot be predicted with certainty and would require further analysis and investigation.

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The reaction rates for trial 1 is 8.22 x 10⁻² M⁻² s⁻¹ and 1.10 M⁻² s⁻¹ for trail 2 and 3

Keep the concentration of W constant while varying the concentrations of U and S while measuring the reaction rate in order to determine the reaction rate with regard to U and S.

Select trial 1 as the reference trial and calculate the reaction's rate constant (k) with respect to U and S, assuming that the concentration of W is constant throughout all three trials.

For trial 1:

[W] = 0.13 M

Rate = 4.72 x 10⁻⁴ M/s

For trial 2:

[W] = 0.13 M

Rate = 1.18 x 10⁻² M/s

From the equation rate = k[U][S], set up the following ratio of rates:

Rate2/Rate1 = (k[U]2[S]2)/(k[U]1[S]1)

Simplifying:

k = (Rate2/Rate1) x (1/[U]2) x (1/[S]2) x ([U]1) x ([S]1)

Substituting the values from trials 1 and 2:

k = (1.18 x 10⁻² M/s) / (4.72 x 10⁻⁴ M/s) x (1/0.65 M) x (1/1 M) x (0.13 M) x (1 M)

k = 8.22 x 10⁻²M⁻² s⁻¹

Similarly, for trial 3:

[W] = 0.13 M

Rate = 2.95 x 10⁻¹ M/s

Again, using trial 1 as the reference trial, figure out the reaction's rate constant (k) in relation to U and S:

k = (Rate3/Rate1) x (1/[U]3) x (1/[S]3) x ([U]1) x ([S]1)

k = (2.95 x 10⁻¹ M/s) / (4.72 x 10⁻⁴ M/s) x (1/3.25 M) x (1/1 M) x (0.13 M) x (1 M)

k = 1.10 M⁻² s⁻¹

Therefore, the equation states the reaction rate in relation to U and S is k = 8.22 x 10⁻² M⁻² s⁻¹ and 1.10 M⁻² s⁻¹ for trials 2 and 3, respectively.

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Answer:its b

Explanation:

jusut bcss

The mass of the silver block, given that it was initially at 55.1 °C  and is submerged into 100.0 g of water at 25.0°C is 189.8 g

We'll begin our calculation by obtaining the heat absorbed by the water. Details below:

Q = MCΔT

Q = 100 × 4.184 × 2.9

Q = 1213.36 J

Finally, we shall determine the mass of the silver block. Details below:

Q = MCΔT

-1213.36 = M × 0.235 × -27.2

-1213.36 = M × -6.392

Divide both sides by -6.392

M = -1213.36 / -6.392

M = 189.8 g

Thus, we can conclude that the mass of the silver block is 189.8 g

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Complete question:

A silver block, initially at 55.1∘C, is submerged into 100.0 g of water at 25.0∘C in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 27.9∘C. The specific heat capacities for water and silver are Cs,water = 4.18J/(g⋅∘C) and Cs, silver = 0.235J/(g⋅∘C). What is the mass of the silver block?

Answer: D

Explanation:

When 12.50 mL of 1.05 M KOH is added to 50.0 mL of 0.225 M HBr, the resulting solution has a pH of 13.35.

Here’s how to calculate it:

First, we need to determine the number of moles of KOH and HBr in the solution:

moles of KOH = (12.50 mL) * (1.05 mol/L) * (1 L/1000 mL) = 0.013125 mol moles of HBr = (50.0 mL) * (0.225 mol/L) * (1 L/1000 mL) = 0.01125 mol

KOH is a strong base and HBr is a strong acid, so they will react completely to form water and a salt (KBr):

KOH + HBr -> KBr + H2O

The number of moles of KOH is greater than the number of moles of HBr, so there will be an excess of KOH in the solution after the reaction is complete:

moles of excess KOH = moles of KOH - moles of HBr = 0.013125 mol - 0.01125 mol = 0.001875 mol

The total volume of the solution is the sum of the volumes of KOH and HBr:

total volume = 12.50 mL + 50.0 mL = 62.5 mL

The concentration of excess OH- ions in the solution is:

[OH-] = moles of excess KOH / total volume = 0.001875 mol / (62.5 mL * (1 L/1000 mL)) = 0.03 M

The pOH of the solution can be calculated using the formula pOH = -log[OH-]:

pOH = -log(0.03) = 1.52

The pH can be calculated using the formula pH + pOH = 14:

pH = 14 - pOH = 14 - 1.52 = 13.35

So the correct answer is D. 13.35.

273.8 grams of zinc (Zn) are needed to produce 8.45 grams of hydrogen gas [tex](H_2).[/tex]

In this chemical reaction, zinc (Zn) reacts with sulfuric acid [tex](H_2SO_4)[/tex] to produce zinc sulfate  [tex](ZnSO_4)[/tex] and hydrogen gas [tex](H_2).[/tex]

From the balanced chemical equation, we can see that 1 mole of zinc (Zn) reacts with 1 mole of sulfuric acid [tex](H_2SO_4)[/tex] to produce 1 mole of hydrogen gas [tex](H_2).[/tex]

Hydrogen gas  has a molar mass of 2.016 g/mol.

Therefore, 8.45 grams of hydrogen gas is equal to 8.45 g / 2.016 g/mol = 4.19 moles of [tex]H_2[/tex].

Since 1 mole of Zn reacts with 1 mole of [tex]H_2[/tex] , we need 4.19 moles of Zn to produce 4.19 moles of [tex]H_2[/tex].

molar mass of Zinc = 65.38 g/mol.

Therefore, 4.19 moles of Zn has a mass of 4.19 moles x 65.38 g/mol = 273.8 grams.

Therefore, you would need 273.8 grams of zinc (Zn) to produce 8.45 grams of hydrogen gas [tex](H_2).[/tex]

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The mass (in grams) of aluminum, Al that must react with iodine to form AlI₃, given that -836.0 KJ of heat is relaesd is 149.0 g

The mass of aluminum required to react with iodine to produce AlI₃ can be obtain as shown below:

2Al(s) + 3I₂(s) → 2AlI₃(s) ∆H = -302.9 KJ

From the balanced equation above,

When -302.9 KJ of heat energy is released, 54 g of aluminum, Al reacted.

Therefore,

When -836.0 KJ of heat energy will be release = (-836.0KJ × 54 g) / -302.9 KJ = 149.0 g of aluminum, Al will react.

Thus, from the above calculation, we can conclude that the mass of aluminum, Al required is 149.0 g

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