Question

What is the percent yield if 107.50 g NH3 reacts with excess O2 according to the
balanced equation below and the actual yield is found to be 154.70 grams of NO?
4 NH3 + 5 02 + 4 NO + 6H2O

Answer 1

Answer:

81.59%

Explanation:

• 4NH₃ + 5O₂ → 4NO + 6H₂O

First we convert 107.50 g of NH₃ into moles, using its molar mass:

• 107.50 g NH₃ ÷ 17 g/mol = 6.32 mol NH₃

Now we calculate how many moles of NO would have been formed by the complete reaction of 6.32 moles of NH₃:

• 6.32 mol NH₃ * $\frac{4molNO}{4molNH_3}$ = 6.32 mol NO

Then we convert 6.32 moles of NO to grams, using its molar mass:

• 6.32 mol NO * 30 g/mol = 189.60 g NO

Finally we calculate the percent yield:

• 154.70 g / 189.60 g * 100% = 81.59%