# If 70.0g nitric acid reacts with 100.0g barium hydroxide what mass of salt may be obtained? If a student collects 138 g of this salt, what is the % yield?

Dec 23, 2015

Theoretical maximum salt mass possible is $145 , 1529 g$.
Actual % yield obtained is 95%.

#### Explanation:

This is an acid-base neutralization/titration reaction and the balanced chemical equation is

$2 H N {O}_{3} + B a {\left(O H\right)}_{2} \to B a {\left(N {O}_{3}\right)}_{2} + 2 {H}_{2} O$

Since a balanced chemical equation represents the mole ratio in which chemicals react, we now first convert all the reactants to moles.

${n}_{H N {O}_{3}} = \frac{m}{{M}_{r}} = \frac{70 g}{\left(1 + 14 + 48\right) g / m o l} = 1 , 111 m o l$.

${n}_{B a {\left(O H\right)}_{2}} = \frac{m}{{M}_{r}} = \frac{100}{137 , 3 + 32 + 2} = 0 , 584 m o l$.

Therefore $B a {\left(O H\right)}_{2}$ is in excess and $H N {O}_{3}$ is the limiting reagent and controls how much of each product is formed.

According to the balanced equation, 2 moles of the acid (nitric acid) produces 1 mole of the salt (barium nitrate).
Hence by ratio and proportion, $1 , 111 m o l$ acid will produce $0 , 5555 m o l$ of the salt.

This will correspond to a mass of salt of
$m = n \times {M}_{r} = 0 , 5555 \times \left(137 , 3 + 28 + 96\right) = 145 , 1529$.

Hence the percentage yield is 138/145.1529xx100=95%.