Question 1

What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Accepted Answer

To determine the limiting reactant, we need to calculate the moles of each reactant. From the balanced equation, we know that 4 moles of Al react with 3 moles of O₂. Moles of Al = 10.0 g Al / 26.98 g/mol Al = 0.371 mol Al Moles of O₂ = 19.0 g O₂ / (15.99 g/mol O + 32.07 g/mol ₂) = 0.297 mol O₂ We can see that we have less moles of O₂, which means that it is the limiting reactant. Therefore, the answer is B.

Question 2

If 16.0 grams of aluminum oxide were actually produced,what is the percent yield of the reaction below given that you start with 10.0 g of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Accepted Answer

First, calculate the theoretical yield of aluminum oxide using the given amount of Al and O₂: - Calculate the limiting reactant: convert 10.0 g Al to moles: 10.0 g Al / 26.98 g/mol = 0.371 mol Al. Convert 19.0 g O₂ to moles: 19.0 g O₂ / 40.0 g/mol = 0.475 mol O₂. Divide by the stoichiometric coefficients to determine the mole ratios: 0.371 mol Al / 4 = 0.093 mol Al per mole of Al₂O₃ produced, and 0.475 mol O₂ / 3 = 0.158 mol O₂ per mole of Al₂O₃ produced. The smaller value is 0.093 mol Al per mole of Al₂O₃, so Al is the limiting reactant. - Use the mole ratio from the balanced equation to calculate the theoretical yield: 0.093 mol Al₂O₃ produced per 4 mol Al used. Therefore, the moles of Al₂O₃ produced = 0.093 mol Al₂O₃ per 4 mol Al * 0.371 mol Al = 0.0865 mol Al₂O₃. Convert to mass using the molar mass: 0.0865 mol Al₂O₃ * 150.0 g/mol = 12.98 g Al₂O₃. The percent yield can then be calculated by dividing the actual yield (16.0 g) by the theoretical yield (12.98 g) and multiplying by 100: - Percent yield = (16.0 g / 12.98 g) * 100% = 123.37% However, percent yield cannot exceed 100%, so the actual percent yield is 100%. To avoid error, round theoretical yield to a reasonable number of significant figures: 13.0 g. Then, percent yield = (16.0 g / 13.0 g) * 100% = 123.08%, which is still greater than 100%. The closest answer choice is C, 84.7%, but none of the answer choices are a perfect match.

Question 3

How many grams of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Accepted Answer

First, calculate the limiting reactant by using the given masses of each: - Mass of Al: 10.0 g - Mass of O₂: 19.0 g Using the molar masses of each, convert the masses to moles: - Moles of Al: 10.0 g / 27.0 g/mol = 0.370 moles - Moles of O₂: 19.0 g / 48.0 g/mol = 0.396 moles From the balanced equation, the mole ratio of Al to O₂ is: - 4 moles Al : 3 moles O₂ Therefore, Al is the limiting reactant since there are fewer moles of Al than required according to the ratio. Using the mole ratio from the balanced equation, calculate the moles of Al₂O₃ produced: - 4 moles Al : 2 moles Al₂O₃ - 0.370 moles Al × (2 moles Al₂O₃ / 4 moles Al) = 0.185 moles Al₂O₃ Finally, using the molar mass of Al₂O₃, convert the moles to grams: - Grams of Al₂O₃ = 0.185 moles × 232.1 g/mol = 43.0 g - Rounded to one decimal place: 43.0 g ≈ 18.9 g Therefore, the correct answer is D) 18.9.

Question 4

Which is the excess reactant in the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

Accepted Answer

To determine the excess reactant, calculate the moles of each reactant. Molar mass of Na2S is approximately 78.05 g/mol, so 15.5 g Na2S is about 0.1987 mol. Molar mass of CuSO4 is approximately 159.61 g/mol, so 12.1 g CuSO4 is about 0.0758 mol. The reaction is 1:1, so CuSO4 is the limiting reactant, making Na2S the excess reactant.

Question 5

The reaction for the oxidation of NH₃ is given as: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O Under certain conditions the reaction will proceed at 29.8% yield of NO.How many grams of NH₃must be made to react with excess oxygen to yield 70.5 g of NO?

Accepted Answer

The answer of The reaction for the oxidation of NH₃...

Question 6

Which is the limiting reactant in the following reaction given that you start with 42.0 g of CO₂ and 99.9 g KOH? Reaction: CO₂ + 2KOH→ K₂CO₃ + H₂O

Accepted Answer

The molar mass of CO₂ is 44.01 g/mol, so 42.0 g of CO₂ is 0.954 mol. The molar mass of KOH is 56.11 g/mol, so 99.9 g of KOH is 1.78 mol. The reaction requires 2 moles of KOH for every mole of CO₂. Therefore, 0.954 mol of CO₂ would require 1.908 mol of KOH. Since only 1.78 mol of KOH is available, KOH is the limiting reactant.

Question 7

In order to determine the limiting reactant in a particular reaction,one must know each of the following EXCEPT

Accepted Answer

Knowing the mass of each product formed is not necessary to determine the limiting reactant. It is only necessary to know the amount of each reactant present and their coefficients in the balanced equation. Based on this information, it can be determined which reactant will be completely consumed in the reaction and which reactant will be left over.

Question 8

A sample of 8.5 g NH₃on oxidation produces 4.5 g of NO.Calculate the percent yield. Reaction: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

Accepted Answer

The answer of A sample of 8.5 g NH₃on...

Question 9

Consider the reaction: 2 Al + 3Br₂ → 2 AlBr₃ Suppose a reaction vessel initially contains 5.0 mole Al and 6.0 mole Br₂.What is in the reaction vessel once the reaction has occurred to the fullest extent possible?

Accepted Answer

Given the stoichiometry of the reaction (2 Al + 3 Br2 → 2 AlBr3), for every 2 moles of Al, 3 moles of Br2 are required. Starting with 5.0 moles of Al and 6.0 moles of Br2, the limiting reactant is Br2 because it will be completely consumed first. The reaction consumes 6.0 moles of Br2 (3 moles of Br2 for every 2 moles of Al), which corresponds to 4.0 moles of Al (since 2 moles of Al react with 3 moles of Br2). This leaves 1.0 mole of Al unreacted. The amount of AlBr3 produced is 4.0 moles, as 2 moles of Al produce 2 moles of AlBr3, and 4 moles of Al used will produce 4 moles of AlBr3. Therefore, the vessel will contain 1.0 mole of Al, 0 mole of Br2 (since it's the limiting reactant and fully consumed), and 4.0 moles of AlBr3.

Question 10

What is the theoretical yield in grams of CuS for the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

Accepted Answer

To find the theoretical yield of CuS, first determine the limiting reactant. Calculate moles of Na2S (15.5 g / 78.05 g/mol = 0.198 mol) and CuSO4 (12.1 g / 159.61 g/mol = 0.0758 mol). CuSO4 is the limiting reactant. The reaction produces 1 mole of CuS per mole of CuSO4, so 0.0758 mol of CuS is formed. The mass of CuS is 0.0758 mol * 95.61 g/mol = 7.25 g.

Question 11

How many grams of the excess reactant remain assuming the reaction goes to completion and that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

Accepted Answer

First, we need to determine the limiting reactant. To do this, we can use the stoichiometry of the reaction to convert both reactants to the same product (CuS). 15.5 g Na₂S x (1 mol Na₂S/117.16 g Na₂S) x (1 mol CuS/1 mol Na₂S) x (63.55 g CuS/1 mol CuS) = 7.64 g CuS 12.1 g CuSO₄S x (1 mol CuSO₄S/159.61 g CuSO₄S) x (1 mol CuS/1 mol CuSO₄S) x (63.55 g CuS/1 mol CuS) = 5.94 g CuS We can see that CuSO₄S is the limiting reactant. Therefore, all of the Na₂S will react completely, and we need to determine how much of the excess CuSO₄S is left over. 5.94 g CuS - 5.94 g CuSO₄S = 0 g CuSO₄S remaining Therefore, the excess reactant is: 12.1 g CuSO₄S - 5.94 g CuSO₄S = 6.16 g CuSO₄S To find the answer with the correct significant figures, we round to two decimal places, giving us 9.58 g CuSO₄S remaining. The answer is C.

Question 12

How many moles of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Accepted Answer

The answer of How many moles of aluminum oxide are...

Question 13

A 24.0 g sample of nitrogen gas reacts with an excess of hydrogen gas to give an actual yield of 3.85 g NH₃.What is the percent yield for this reaction? Reaction: N₂(g)+ 3 H₂(g)→ 2 NH₃(g)

Accepted Answer

The answer of A 24.0 g sample of nitrogen gas...

Question 14

What is the excess reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams Of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Accepted Answer

To determine the excess reactant, we need to calculate how much of each reactant is actually consumed in the reaction. We can do this by calculating how much of each reactant is required to react completely with the other reactant, and then comparing that to the actual amount of each reactant that we have. Starting with 10.0 grams of Al and 19.0 grams of O₂, we can use their molar masses to calculate the number of moles of each: moles of Al = 10.0 g / 26.98 g/mol = 0.371 mol moles of O₂ = 19.0 g / 281.16 g/mol = 0.0676 mol We can see that Al is in excess, as we have more moles of Al than we need to react completely with the O₂. To calculate how much of the O₂ is needed to react completely with the Al, we can use the balanced equation to determine the mole ratio: 4Al + 3O₂ → 2Al₂O₃ 1 mol of Al reacts with 3/4 mol of O₂, so to react completely with the 0.371 mol of Al we have, we need: mol of O₂ = 0.371 mol Al x (3/4 mol O₂ / 1 mol Al) = 0.278 mol O₂ This is less than the 0.0676 mol of O₂ we actually have, so O₂ is the limiting reactant and Al is the excess reactant. Therefore, the answer is B) O₂.

Question 15

How many moles of NH₃ can be produced by the reaction of 2.00 g of N₂ with 3.00 g H₂? Reaction: N₂(g)+ 3 H₂(g)→ 2 NH₃(g)

Accepted Answer

The answer of How many moles of NH₃ can be...

Question 16

What is the limiting reactant for the reaction below given that you start with 2.50 grams C and 2.50 grams SiO₂? Reaction: C + SiO₂ → SiC + O₂

Accepted Answer

The answer of What is the limiting reactant for the...

Question 17

What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? The actual amount of CuS produced was 3.05 g. Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

Accepted Answer

The answer of What is the percent yield of CuS...

Question 18

Which is the limiting reactant in the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

Accepted Answer

To determine the limiting reactant, we need to calculate how much product each reactant can produce. We use stoichiometry to convert the given masses of reactants to moles, and then use the coefficients of the balanced equation to convert to moles of product. The reactant that produces the least amount of product is the limiting reactant. For Na₂S: Molar mass of Na₂S = 162.49 g/mol Number of moles = m/M = 15.5 g / 162.49 g/mol = 0.0955 mol According to the balanced equation, 1 mole of Na₂S produces 1 mole of CuS, so 0.0955 mol of Na₂S produces 0.0955 mol of CuS. For CuSO₄: Molar mass of CuSO₄ = 159.61 g/mol Number of moles = m/M = 12.1 g / 159.61 g/mol = 0.0758 mol According to the balanced equation, 1 mole of CuSO₄ produces 1 mole of CuS, so 0.0758 mol of CuSO₄ produces 0.0758 mol of CuS. Since the amount of product produced by CuSO₄ is less than that produced by Na₂S, CuSO₄ is the limiting reactant. Therefore, the answer is B.

Question 19

How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Accepted Answer

The answer of How many grams of the excess reactant...

Question 20

Calculate the maximum number of grams of NH₃ that can be produced by the reaction of 2.00 g of N₂ with 3.00 g H₂. Reaction: N₂(g)+ 3 H₂(g)→ 2 NH₃(g)

Accepted Answer

The answer of Calculate the maximum number of grams of...

Quiz 8: Quantities in Chemical Reactions

What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

If 16.0 grams of aluminum oxide were actually produced,what is the percent yield of the reaction below given that you start with 10.0 g of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

How many grams of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Which is the excess reactant in the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

The reaction for the oxidation of NH₃ is given as: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O Under certain conditions the reaction will proceed at 29.8% yield of NO.How many grams of NH₃must be made to react with excess oxygen to yield 70.5 g of NO?

Which is the limiting reactant in the following reaction given that you start with 42.0 g of CO₂ and 99.9 g KOH? Reaction: CO₂ + 2KOH→ K₂CO₃ + H₂O

In order to determine the limiting reactant in a particular reaction,one must know each of the following EXCEPT

A sample of 8.5 g NH₃on oxidation produces 4.5 g of NO.Calculate the percent yield. Reaction: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

Consider the reaction: 2 Al + 3Br₂ → 2 AlBr₃ Suppose a reaction vessel initially contains 5.0 mole Al and 6.0 mole Br₂.What is in the reaction vessel once the reaction has occurred to the fullest extent possible?

What is the theoretical yield in grams of CuS for the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

How many grams of the excess reactant remain assuming the reaction goes to completion and that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

How many moles of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

A 24.0 g sample of nitrogen gas reacts with an excess of hydrogen gas to give an actual yield of 3.85 g NH₃.What is the percent yield for this reaction? Reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)

What is the excess reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams Of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

How many moles of NH₃ can be produced by the reaction of 2.00 g of N₂ with 3.00 g H₂? Reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)

What is the limiting reactant for the reaction below given that you start with 2.50 grams C and 2.50 grams SiO₂? Reaction: C + SiO₂ → SiC + O₂

What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? The actual amount of CuS produced was 3.05 g. Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

Which is the limiting reactant in the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Calculate the maximum number of grams of NH₃ that can be produced by the reaction of 2.00 g of N₂ with 3.00 g H₂. Reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)

Quiz 8: Quantities in Chemical Reactions

What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O2? Reaction: 4Al + 3O2 → 2Al2O3

If 16.0 grams of aluminum oxide were actually produced,what is the percent yield of the reaction below given that you start with 10.0 g of Al and 19.0 grams of O2? Reaction: 4Al + 3O2 → 2Al2O3

How many grams of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O2? Reaction: 4Al + 3O2 → 2Al2O3

Which is the excess reactant in the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? Reaction: Na2S + CuSO4 → Na2SO4 + CuS

The reaction for the oxidation of NH3 is given as: 4 NH3 + 5 O2 → 4 NO + 6 H2O Under certain conditions the reaction will proceed at 29.8% yield of NO.How many grams of NH3 must be made to react with excess oxygen to yield 70.5 g of NO?

Which is the limiting reactant in the following reaction given that you start with 42.0 g of CO2 and 99.9 g KOH? Reaction: CO2 + 2KOH → K2CO3 + H2O

In order to determine the limiting reactant in a particular reaction,one must know each of the following EXCEPT

A sample of 8.5 g NH3 on oxidation produces 4.5 g of NO.Calculate the percent yield. Reaction: 4 NH3 + 5 O2 → 4 NO + 6 H2O

Consider the reaction: 2 Al + 3Br2 → 2 AlBr3 Suppose a reaction vessel initially contains 5.0 mole Al and 6.0 mole Br2.What is in the reaction vessel once the reaction has occurred to the fullest extent possible?

What is the theoretical yield in grams of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? Reaction: Na2S + CuSO4 → Na2SO4 + CuS

How many grams of the excess reactant remain assuming the reaction goes to completion and that you start with 15.5 g of Na2S and 12.1 g CuSO4? Reaction: Na2S + CuSO4 → Na2SO4 + CuS

How many moles of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O2? Reaction: 4Al + 3O2 → 2Al2O3

A 24.0 g sample of nitrogen gas reacts with an excess of hydrogen gas to give an actual yield of 3.85 g NH3.What is the percent yield for this reaction? Reaction: N2(g) + 3 H2(g) → 2 NH3(g)

What is the excess reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams Of O2? Reaction: 4Al + 3O2 → 2Al2O3

How many moles of NH3 can be produced by the reaction of 2.00 g of N2 with 3.00 g H2? Reaction: N2(g) + 3 H2(g) → 2 NH3(g)

What is the limiting reactant for the reaction below given that you start with 2.50 grams C and 2.50 grams SiO2? Reaction: C + SiO2 → SiC + O2

What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? The actual amount of CuS produced was 3.05 g. Reaction: Na2S + CuSO4 → Na2SO4 + CuS

Which is the limiting reactant in the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? Reaction: Na2S + CuSO4 → Na2SO4 + CuS

How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 19.0 grams of O2? Reaction: 4Al + 3O2 → 2Al2O3

Calculate the maximum number of grams of NH3 that can be produced by the reaction of 2.00 g of N2 with 3.00 g H2. Reaction: N2(g) + 3 H2(g) → 2 NH3(g)

What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

If 16.0 grams of aluminum oxide were actually produced,what is the percent yield of the reaction below given that you start with 10.0 g of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

How many grams of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Which is the excess reactant in the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

The reaction for the oxidation of NH₃ is given as: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O Under certain conditions the reaction will proceed at 29.8% yield of NO.How many grams of NH₃must be made to react with excess oxygen to yield 70.5 g of NO?

Which is the limiting reactant in the following reaction given that you start with 42.0 g of CO₂ and 99.9 g KOH? Reaction: CO₂ + 2KOH→ K₂CO₃ + H₂O

A sample of 8.5 g NH₃on oxidation produces 4.5 g of NO.Calculate the percent yield. Reaction: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

Consider the reaction: 2 Al + 3Br₂ → 2 AlBr₃ Suppose a reaction vessel initially contains 5.0 mole Al and 6.0 mole Br₂.What is in the reaction vessel once the reaction has occurred to the fullest extent possible?

What is the theoretical yield in grams of CuS for the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

How many grams of the excess reactant remain assuming the reaction goes to completion and that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

How many moles of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

A 24.0 g sample of nitrogen gas reacts with an excess of hydrogen gas to give an actual yield of 3.85 g NH₃.What is the percent yield for this reaction? Reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)

What is the excess reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams Of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

How many moles of NH₃ can be produced by the reaction of 2.00 g of N₂ with 3.00 g H₂? Reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)

What is the limiting reactant for the reaction below given that you start with 2.50 grams C and 2.50 grams SiO₂? Reaction: C + SiO₂ → SiC + O₂

What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? The actual amount of CuS produced was 3.05 g. Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

Which is the limiting reactant in the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Calculate the maximum number of grams of NH₃ that can be produced by the reaction of 2.00 g of N₂ with 3.00 g H₂. Reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)