How many liters of a 3.0 M H3PO4 solution are required to react with 4.5 g of zinc? __ H3PO4 + __ Zn _x0001_ __ Zn3(PO4)2 + __ H2

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##