First, calculate the molar mass of aluminum (Al):
M(Al) = 4.5 g / mol

From the balanced chemical equation:
2 moles of Al produce 3 moles of H2

Using the molar ratio, we can calculate the moles of H2 produced when 4.5g of Al reacts:
Moles of Al = 4.5 g / M(Al) = 4.5 g / 27 g/mol ≈ 0.1667 mol

Therefore, according to the molar ratio:
Moles of H2 = 3/2 Moles of Al = 3/2 0.1667 mol ≈ 0.25 mol

Now, using the molar volume of gases at standard temperature and pressure (STP):
1 mol of any ideal gas occupies approximately 22.4 L at STP

Thus, the volume of H2 produced:
V(H2) = 0.25 mol * 22.4 L/mol ≈ 5.6 L

Therefore, when 4.5g of Al reacts, the volume of H2 produced is approximately 5.6 liters.