Question

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

4NH₃ +5O₂ → 4NO + 6H₂O 

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

Answer 1

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.

Answer 2

Answer 3

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.