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den301095 [7]
3 years ago
15

What is the percent yield if 107.50 g NH3 reacts with excess O2 according to the

Chemistry
1 answer:
Maksim231197 [3]3 years ago
8 0

Answer:

81.59%

Explanation:

  • 4NH₃ + 5O₂ → 4NO + 6H₂O

First we <u>convert 107.50 g of NH₃ into moles</u>, using its <em>molar mass</em>:

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

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

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

Then we <u>convert 6.32 moles of NO to grams</u>, using its <em>molar mass</em>:

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

Finally we <u>calculate the percent yield</u>:

  • 154.70 g / 189.60 g * 100% = 81.59%

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