Question

How many grams of dihydrogen gas are required to produce a `10.53*g` mass of ammonia?

Answer 1

Almost `2*g` of dihydrogen gas are required.

`1/2N_2(g) + 3/2H_2(g) rarr NH_3(g)`

Explanation:

We follow the given stoichiometric equation, which explicitly states that `14*g` dinitrogen gas react with `3.0*g` dihydrogen to give `17*g` ammonia.

`"Moles of ammonia"` `=` `(10.53*g)/(17.03*g*mol^-1)` `=` `0.618*mol`

And thus `3/2xx0.618*molxx2.0159*g*mol^-1" dihydrogen"` are required.