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What is the theoretical yield of water, in moles, when 40.0

Homewroklib Q&A #1267 - What is the theoretical yield of water, in moles, when 40.0

What is the theoretical yield of water, in moles, when 40.0

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

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Homeworklib: Your Answer

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.
 

Homeworklib: Your Answer #2

What is the theoretical yield of water, in moles, when 40.0

Homeworklib: Your 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.

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