Question

How many liters of a 3.0M H3(PO4) solution are required to react with 4.5 g of zinc?

Answer 1

`0.015"L H"_3("PO"_4)`

Explanation:

1. Write a balanced equation
   `3Zn(s)+2H_3(PO_4)(aq)->Zn_3(PO_4)_2(s)+ 3H_2(g)`

2. Stoichiometry
   `4.5 cancel("g Zn")* (1cancel("mol Zn"))/ (65.38 cancel("g Zn")) * (2 cancel("mol H"_3("PO"_4) ))/ (3 cancel("mol Zn"))* (1 "L H"_3("PO"_4))/ (3 cancel("mol H"_3("PO"_4)))=`

`0.015"L H"_3("PO"_4)`