Question

What mass of `"ammonium sulfate"` can be prepared from a `30*g` mass of `"ammonia"`, and a `196*g` mass of `"sulfuric acid"`?

Answer 1

`2NH_3(aq) + H_2SO_4(aq) rarr (NH_4)_2SO_4(aq)`

Explanation:

We have a balanced chemical equation that represents the neutralization of ammonia by sulfuric acid.

`"Moles of ammonia"` `=` `(30*g)/(17*g*mol^-1)` `=` `1.76*mol`.

`"Moles of vitriol"` `=` `(196*g)/(98.08*g*mol^-1)` `=` `2.00*mol`.

Now clearly, given the 2:1 stoichiometry, ammonia is the limiting reagent, and sulfuric is in excess. (What do I mean by 2:1 stoichiometry?)

Given `1.76` `mol` ammonia, we can make `(1.76*mol)/2` ammonium sulfate;

i.e. `(1.76*cancel(mol))/2xx132.1*g*cancel(mol^-1)` `=` `??g`