The action of some commercial drain cleaners is based on the following reaction:2 NaOH(s) + 2 Al(s) + 6 H2O(l) → 2 NaAl(OH)4(s) + 3 H2(g)

Question

What is the volume of H2 gas formed at STP when 4.32 g of Al reacts with excess NaOH?

~the answer says its 5.38 L but im trying to understand how to get that. Thank you in advance:)

Answer 1

From the balanced equation, we see that the molar ratio of H₂ to Al is 3:2. Since the problem states that the system is at STP, we know that there are 22.4 L per mol of gas. So, in setting up the stoichiometry we have:

(22.4 L H₂/1 mol H₂) (3 mol H₂/2 mol Al) (1 mol Al/26.98 g Al) 4.32 g Al = 5.38 L H₂ (to 3 sig figs)