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?

A. 9.12 L

B. 9.95 L

C. 6.08 L

D. 4.05 L

The volume of H₂ gas formed: V = 7.87 L

Explanation:

Given reaction: 2 NaOH (s) + 2 Al (s) + 6 H₂O (l) → 2 NaAl (OH) ₄ (s) + 3 H₂ (g)

In the given reaction, number of moles of Al = 1, number of moles of H₂ = 3

Given mass of Al = 6.32 g, atomic mass of Al = 26.98 g/mol, atomic mass of H₂ gas = 2.015 g/mol

∴ Number of moles of Al = given mass : atomic mass = 6.32 g : 26.98 g/mol = 0.234 mol

In the given reaction, 3 moles hydrogen gas (H₂) is obtained when 2 moles aluminum (Al) reacts with excess sodium hydroxide (NaOH)

Therefore, number of moles of hydrogen gas (H₂) obtained when 0.234 mol Al reacts with excess NaOH = 0.234 mol * 3 mol : 2 mol = 0.351 mol

Given: Standard temperature: T = 273.15 K, Standard pressure: P = 1 atm,

Number of moles of H₂ gas: n = 0.351 mol, gas constant: R = 0.08206 L·atm / (mol·K)

Volume of H₂ gas: V = ?

According to the ideal gas equation: P·V = n·R·T

⇒ V = n·R·T : P

⇒ V = (0.351 mol) * (0.08206 L·atm / (mol·K)) * (273.15 K) : (1 atm)

⇒ V = 7.87 L

Therefore, the volume of H₂ gas formed: V = 7.87 L