Given the equation:$N_2+3H_2 rarr 2NH_3$, what volume of $NH_3$ at STP is produced if 25 grams of $N_2$ is reacted with an excess of $H_2$?

Md Ariful Islam

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Md Ariful Islam

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One thing I think it's interesting to note is that you'll never be able to have that reaction happen in the standard temperature and pressure; you can, however, have the resulting ammonia in the STP.

Anyhow, $N_2$ has molar mass $ 14*2=28$, so for 25 g we have

$n_(N_2) = m/(MM) = 25/28 mol$

Every one mol of nitrogen makes two moles of ammonia, so we have

$n_(NH_3) = 2n_(N_2) = 2*25/28 = 25/14 mol$

Since we know that at STP 1 mol of gas has 22.4 L (this can easily be checked by plugging the appropriate values in $pV = nRT$) We have

$V_(NH_3) = 22.4*25/14 = 40 L$

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