In the equation $N_2(g) + 3H_2(g) -> 2NH_3(g)$, 8.5 grams of hydrogen and 15.8 grams of nitrogen react to form ammonia. How many grams of ammonia are produced?

Md Ariful Islam

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Md Ariful Islam

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Moles of $N_2$ reactant $=$ $(15.8*g)/(28.02*g*mol^-1)$ $=$ $0.564$ $mol$.

Moles of $H_2$ reactant $=$ $(8.5*g)/(2.02*g*mol^-1)$ $=$ $4.21$ $mol$.

Dihydrogen is the reagent in excess (clearly!). Given that dinitrogen is the limiting reagent, at most I can form $1.13$ mol ammonia gas (i.e. 2 equiv with respect to dinitrogen).

If I have formed $1.13$ $mol$ ammonia, this is $1.13$ $cancel("mol")xx17.0*g*cancel("mol"^-1)$ $=$ $??g$

You might already know that this is probably the most important reaction on the planet. An industrialist would have an orgasm if he could achieve such yields.

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