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

1 Answer
Stefan V.
Dec 6, 2014

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

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