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

Sep 4, 2016

$2 N {H}_{3} \left(a q\right) + {H}_{2} S {O}_{4} \left(a q\right) \rightarrow {\left(N {H}_{4}\right)}_{2} S {O}_{4} \left(a q\right)$

#### Explanation:

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

$\text{Moles of ammonia}$ $=$ $\frac{30 \cdot g}{17 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.76 \cdot m o l$.

$\text{Moles of vitriol}$ $=$ $\frac{196 \cdot g}{98.08 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.00 \cdot m o l$.

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 $1.76$ $m o l$ ammonia, we can make $\frac{1.76 \cdot m o l}{2}$ ammonium sulfate;

i.e. $\frac{1.76 \cdot \cancel{m o l}}{2} \times 132.1 \cdot g \cdot \cancel{m o {l}^{-} 1}$ $=$ ??g