2Al(s) + 6HCL (aq) = 2 AlCl3 (aq) + 3H2 (g)

Question

What volume of H2(g) is produced when 8.60 g Al(s) reacts at STP?

They want it in L

Answer 1

At STP, 1 mole of any ideal gas = 22.4 liters (L)

moles Al present = 8.60 g x 1 mol Al/26.98 g = 0.319 moles Al

moles H2 produced = 0.319 moles Al x 3 moles H2 / 2 moles Al = 0.4785 moles H2 produced

liters of H2 = 0.4785 moles x 22.4 L/mole = 10.7 L (to 3 signifiant figures)