Respuesta :
Answer:
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
