# Question 79474

Oct 6, 2016

A bit under $5 \cdot g$ of ammonia will be produced.

#### Explanation:

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

$\text{Moles of } C a {\left(O H\right)}_{2}$ $=$ $\frac{10.0 \cdot g}{74.09 \cdot g \cdot m o {l}^{-} 1} = 0.135 \cdot m o l$

And thus, given the stoichiometry, $0.270 \cdot m o l$ of ammonia will be generated.

This has a mass of 0.270*molxx17.03*g*mol^-1=??g#