How many moles of nitrogen would react with excess hydrogen to produce 520 mL of ammonia?

0.0116mol

0.012mol

0.0232mol

0.024mol

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14 November, 22:47

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Deacon Williamson · Answer 1

N₂ + 3H₂ = 2NH₃

n (N₂) = V (NH₃) / {2Vm}

V (NH₃) = 520 mL=0.520L

Vm=22.4 L/mol

n (N₂) = 0.520/{2*22.4}=0.0116 mol

How many moles of nitrogen would react with excess hydrogen to produce 520 mL of ammonia?0.0116mol0.012mol0.0232mol0.024mol