# How many liters of a 3.0 M H3PO4 solution are required to react with 4.5 g of zinc? __ H3PO4 + __ Zn  __ Zn3(PO4)2 + __ H2

Dec 6, 2014

The volume is $15.6 m L$, or $0.0156 L$

$2 {H}_{3} P {O}_{4} + 3 Z n \to Z {n}_{3} {\left(P {O}_{4}\right)}_{2} + 3 {H}_{2}$

As you can see, we have a $2 : 3$ mole ratio between ${H}_{3} P {O}_{4}$ and $Z n$. Knowing that the molar mass of $Z n$ is $65.4 \frac{g}{m o l}$, we can determine the number of $Z n$ moles to be

${n}_{Z n} = \frac{m}{m o l a r m a s s} = \frac{4.5 g}{65.4 \frac{g}{m o l}} = 0.07$

This means that the number of ${H}_{3} P {O}_{4}$ moles is equal to

${n}_{{H}_{3} P {O}_{4}} = 0.07 \cdot \frac{2}{3} = 0.047$

Therefore, the volume required is

$V = {n}_{{H}_{3} P {O}_{4}} / C = \frac{0.047 m o l e s}{3 \frac{m o l e s}{L}} = 0.0156 L = 15.6 m L$