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Today, 08:06

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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  1. Today, 08:22
    0
    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
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