how many liters of a 3.0 M H3PO4 solition are required to react with 4.5 g of zinc?

Molar mass of Zn = 65.4 g/mol
No. of moles of Zn reacted = (4.5 g) / (65.4 g/mol) = 0.06881 mol

3Zn + 2H₃PO₄ → Zn₃(PO₄)₂ + 3H₂
Mole ratio Zn : H₃PO₄ = 3 : 2
No. of moles of H₃PO₄ required = (0.06881 mol) × (2/3) = 0.04587 mol

Volume of H₃PO₄ required = (0.04587 mol) / (0.3 mol/L) = 0.153 L

First write down the BALANCED reaction equation.
2H3PO4 + 3Zn = (Zn)3(PO4)2 + 3H2
The molar ratios are 2:3::1:3
Next calculate the moles of zinc
mol(Zn) = 4.5g / 65 = 0.06923 moles ( Equivalent to '3' in the molar ratios).
mol(H3PO4) = 0.06923 x 2/3 = 0.04615 moles (Equivalent to '2' in the molar ratios).
Next using the equation
moles = [conc] x vol(mL) / 1000
Algeraically rearrange
vol(mL) / 1000 = moles / [Conc]
vol(mL) / 1000 = 0.04615 / 3.0
vol(L) = 0.0153 L