Question

Aqueous sulfuric acid H

2

SO

4

 reacts with solid sodium hydroxide NaOH

 to produce aqueous sodium sulfate Na

2

SO

4

 and liquid water H

2

O

. If 9.14

g

 of water is produced from the reaction of 85.3

g

 of sulfuric acid and 36.9

g

 of sodium hydroxide, calculate the percent yield of water.

Answer 1

Write a balanced chemical equation;

$H_2SO_4+2NaOH \to Na_2SO_4+2H_2O$

moles of $H_2SO_4$ $=85.3/98$

moles of $NaOH=36.9/40$

$NaOH$ has less moles hence it is the limiting reagent.

Now 2moles of $NaOH$ = 2moles of $H_2O$ formed.

Hence moles of water formed $=36.9/40$

Mass of water produced (theoretical) $=\frac{36.9}{40}\times 18=16.605g$

Actual mass of water produced $=9.14g$

$\%yield=\frac{9.14}{16.605}\times 100\%=55.04\%$