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

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.