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

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answer; 4 electrons;