A 2.0 g sample of an alloy containing Al and Mg is treated with excess HCl to produce 2.43 L of H 2 gas at 2 7 ∘ C and 1 atm . Another 2.0 g sample of the same alloy is treated with excess NaOH solu

Question

A 2.0 g sample of an alloy containing Al and Mg is treated with excess HCl to produce 2.43 L of $H_{2}$ gas at $2 7^{\circ} C$ and 1 atm . Another 2.0 g sample of the same alloy is treated with excess NaOH solution to produce 1.66 L of $H_{2}$ gas under the same conditions.

What is the mass fraction of Mg in the alloy?

(Atomic masses: Al=27, Mg=24; R=0.0821 L atm K^-1 mol^-1 )

TestHub · Accepted Answer

Explain the question:

Find the mass fraction of Mg in an Al–Mg alloy using its gas evolution reactions with HCl and NaOH.

Concept:

Ideal Gas Law: $P V = n RT$ to find moles of H₂ gas produced.

Solution:

From the reaction with NaOH, only Aluminum reacts to produce H₂ gas according to the equation:

$2 A l + 2 N a O H + 6 H_{2} O \to 2 N a [A l (O H)_{4}] + 3 H_{2}$

First, calculate the moles of H₂ produced in this reaction at $T = 27° C = 300 K$ :

$n_{H_{2}, N a O H} = \frac{P \times V}{R \times T} = \frac{1 a t m \times 1.66 L}{0.0821 L a t m K ^{- 1} m o l ^{- 1} \times 300 K} = \frac{1.66}{24.63} \approx 0.0674 m o l$

From stoichiometry, 1.5 moles of H₂ are produced per mole of Al, so:

Moles of Al $= \frac{0.0674 m o l}{1.5} \approx 0.0449 m o l$

Mass of Al $= 0.0449 m o l \times 27 g / m o l \approx 1.213 g$

Since the total mass of the alloy sample is 2.0 g, the mass of Magnesium is:

Mass of Mg $= 2.0 g - 1.213 g = 0.787 g$

Finally, calculate the mass fraction of Magnesium:

Mass fraction of Mg $= \frac{0.787 g}{2.0 g} = 0.3935 \approx 0.40$

This result is consistent with the HCl data where 0.787 g Mg (0.0328 mol) and 1.213 g Al (gives 0.0674 mol H₂) together produce $\approx 0.100 m o l H_{2}$ , which occupies $\approx 2.46 L$ at the given conditions. Final Answer: Option (A)

Chemistry - Mole Concept Question with Solution | TestHub

Options:

Doubts & Discussion