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 8.64 g of Al reacts with excess NaOH?

Question

Quizplus · Accepted Answer

First, we need to balance the equation:
2 NaOH(s) + 2 Al(s) + 6 H2SO4(l) → 2 NaAl(OH)4(s) + 3 H2S(g)

Now we can use stoichiometry to find the moles of H2S produced:
8.64 g Al × (1 mol Al/ 26.98 g Al) × (3 mol H2S / 2 mol Al) = 1.01 mol H2S

At STP, 1 mole of gas occupies 22.4 L. Therefore, the volume of H2S at STP is:
1.01 mol H2S × 22.4 L/mol = 22.6 L

However, this is the volume at STP, not the actual conditions present in the drain. We don't know the temperature and pressure of the drain, so we assume STP for simplicity.

Therefore, the answer is:
C) 10.8 L

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