Question

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?

Answer 1

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

`2HNO_3+Ba(OH)_2->Ba(NO_3)_2+2H_2O`

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

`n_(HNO_3)=m/(M_r)=(70g)/((1+14+48)g//mol)=1,111 mol`.

`n_(Ba(OH)_2)=m/(M_r)=100/(137,3+32+2)=0,584 mol`.

Therefore `Ba(OH)_2` is in excess and `HNO_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 mol` acid will produce `0,5555 mol` of the salt.

This will correspond to a mass of salt of
`m=nxxM_r=0,5555xx(137,3+28+96)=145,1529`.

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