Question

# Ammonia reacts with oxygen to form nitric oxide and water vapor: 4NH3 +5O2 → 4NO + 6H2O

Ammonia reacts with oxygen to form nitric oxide and water vapor:

4NH3 +5O2 → 4NO + 6H2

What is the theoretical yield of water, in moles, when 40.0g NH3 and 50.0g O2 are mixed and allowed to react:

•  1.87 mol

•  3.53 mol

•  1.57 mol

•  1.30 mol

•  None of these

Number of moles of NH3= mass/mol. wt .=40.0 g/ (17.03 g/mol)=2.348796 mol

Number of moles of O2= mass / mol . wt .=50.0 g /(17.03 g / mol)=1.5625 mol

According to balanced equation,

5 moles of O2 reacts with 4 moles of NH3

So, 1.5625 mol of O2 would react with 1.5625 mol * 4 / 5 = 1.25 mol of NH3

But the number of moles of NH3 present is more than required.

That is NH3 is the excess reagent.

O2 is the Limiting reagent.

According to the balanced equation,

5 moles of O2 produces 6 moles of H2O

So, 1.5625 mol of O2 would produce 1.5625 mol * 6 / 5=1.875 mol of H2O

The number of moles of H2O = 1.87 mol

Option A is the correct answer.

To calculate the theoretical yield of water (H2O) in the given reaction, we first need to identify the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction, thereby limiting the amount of products that can be formed.

Let's first calculate the number of moles of NH3 and O2 present in the given reaction:

Number of moles of NH3 = 40.0 g / (17.03 g/mol) = 2.35 mol

Number of moles of O2 = 50.0 g / (32.00 g/mol) = 1.56 mol

According to the balanced chemical equation, 4 moles of NH3 react with 5 moles of O2 to produce 6 moles of H2O. Therefore, if NH3 is the limiting reactant, the theoretical yield of water can be calculated as:

Theoretical yield of H2O = (2.35 mol NH3) x (6 mol H2O / 4 mol NH3) = 3.53 mol H2O

Similarly, if O2 is the limiting reactant, the theoretical yield of water can be calculated as:

Theoretical yield of H2O = (1.56 mol O2) x (6 mol H2O / 5 mol O2) = 1.87 mol H2O

Since the calculated theoretical yield of H2O using NH3 as the limiting reactant is higher than that using O2 as the limiting reactant, NH3 is the limiting reactant in this case. Therefore, the theoretical yield of water is 3.53 mol.

Hence, the correct answer is 3.53 mol.

1. To find the theoretical yield of water (H2O) in moles, we need to follow the stoichiometry of the balanced equation:

2. 4NH3 + 5O2 → 4NO + 6H2O

3. First, let's calculate the number of moles for NH3 and O2 using their respective masses and molar masses:

4. Molar mass of NH3 (Ammonia) = 14.01 g/mol + 3(1.01 g/mol) = 17.03 g/mol

5. Moles of NH3 = mass of NH3 / molar mass of NH3 = 40.0 g / 17.03 g/mol ≈ 2.35 mol

6. Molar mass of O2 (Oxygen) = 2(16.00 g/mol) = 32.00 g/mol

7. Moles of O2 = mass of O2 / molar mass of O2 = 50.0 g / 32.00 g/mol ≈ 1.56 mol

8. Now, we can determine the limiting reactant by comparing the moles of NH3 and O2. The reactant that produces the least amount of product is the limiting reactant.

9. From the balanced equation, we can see that the stoichiometric ratio between NH3 and H2O is 4:6 (4 moles of NH3 produce 6 moles of H2O).

10. Moles of H2O formed from NH3 = 2.35 mol NH3 × (6 mol H2O / 4 mol NH3) ≈ 3.52 mol H2O

11. From the balanced equation, the stoichiometric ratio between O2 and H2O is 5:6 (5 moles of O2 produce 6 moles of H2O).

12. Moles of H2O formed from O2 = 1.56 mol O2 × (6 mol H2O / 5 mol O2) ≈ 1.87 mol H2O

13. Since the theoretical yield is limited by the reactant that produces the least amount of product, the limiting reactant is O2, and the theoretical yield of water (H2O) is approximately 1.87 moles.

source: me.

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