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)
What is the volume of H₂ gas formed at STP when 6.32 g of Al reacts with excess NaOH?

Question

Quizplus · Accepted Answer

First, we need to use stoichiometry to find the amount of H2S gas that is produced from 6.32 g of Al:

1 mol Al = 26.98 g Al
2 mol NaOH = 2 × 40.00 g NaOH = 80.00 g NaOH
6 mol H2S = 6 × 34.08 g H2S = 204.48 g H2S

6.32 g Al × (1 mol Al / 26.98 g Al) × (6 mol H2S / 2 mol Al) × (22.4 L H2S / 1 mol H2S) = 7.87 L H2S gas at STP

(Note: We used the conversion factor of 22.4 L/mol for gases at STP, which is the molar volume.)

Therefore, the answer is C) 7.87 L.

The Action of Some Commercial Drain Cleaners Is Based on the Following