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

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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.


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