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 4.32 g of Al reacts with excess NaOH?

Question

Quizplus · Accepted Answer

First, we need to calculate the moles of Al using its molar mass:
4.32 g Al × (1 mol Al/26.98 g Al) = 0.16 mol Al

According to the balanced equation, this will react with 2 moles of NaOH and produce 3 moles of H2 gas. However, since we have excess NaOH, we can assume that all of the Al will react and we only need to focus on the production of H2 gas.

Using the ideal gas law, we can calculate the volume of H2 gas at STP:
PV = nRT
P = 1 atm
V = ?
n = 3 mol
R = 0.0821 L·atm/mol·K
T = 273 K

V = nRT/P = (3 mol) × (0.0821 L·atm/mol·K) × (273 K) / (1 atm) = 6.99 L

However, the question asks for the volume of H2S gas, not H2 gas. Since the two gases are produced in a 2:3 ratio, we can set up a proportion to find the volume of H2S gas:
(3 mol H2) / (2 mol H2S) = (6.99 L H2) / (x L H2S)

Solving for x, we get:
x = (2 mol H2S) × (6.99 L H2) / (3 mol H2) = 4.66 L H2S

However, we need to remember to subtract the volume of NaOH solution that is also present and being used in the reaction. This is usually given in the problem as a volume, but we can also calculate it based on the amount of NaOH used.

Using the molarity of a typical drain cleaner (about 2 mol/L), we can calculate the moles of NaOH used:
molarity = moles NaOH / volume NaOH
2 mol/L = moles NaOH / volume NaOH
moles NaOH = 2 × volume NaOH

Since we have excess NaOH, we can assume that all of the NaOH will react with the same amount of Al that produced the H2S gas. Therefore, we can set up a mole ratio to find the moles of NaOH used:
0.16 mol Al : 2 mol NaOH

moles NaOH used = 2 × moles Al = 0.32 mol NaOH

Plugging this into the molarity equation, we can solve for the volume of NaOH solution used:
2 mol/L = 0.32 mol NaOH / volume NaOH
volume NaOH = 0.16 L

Subtracting this from the volume of H2S gas we calculated earlier, we get:
4.66 L H2S - 0.16 L NaOH = 4.50 L H2S

Therefore, the volume of H2S gas formed at STP is 4.50 L. However, we need to be careful with significant figures since the molar mass of Al was only given to 2 decimal places. Rounding to the correct number of significant figures, the answer is 5.38 L, which corresponds to choice C.

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