Question

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

Answer 1

The volume is `15.6 mL`, or `0.0156 L`

Start with the balanced equation:

`2H_3PO_4 + 3Zn -> Zn_3(PO_4)_2 + 3H_2`

As you can see, we have a `2:3` mole ratio between `H_3PO_4` and `Zn`. Knowing that the molar mass of `Zn` is `65.4 g/(mol)`, we can determine the number of `Zn` moles to be

`n_(Zn) = m/(molar mass) = (4.5 g)/(65.4 g/(mol)) = 0.07`

This means that the number of `H_3PO_4` moles is equal to

`n_(H_3PO_4) = 0.07 * 2/3 = 0.047`

Therefore, the volume required is

`V = n_(H_3PO_4)/C = (0.047 m ol e s)/(3 (mo l e s)/L) = 0.0156 L = 15.6 mL`