A 5.34 g sample of an aluminum alloy is reacted with hydrochloric acid to produce aluminum chloride and hydrogen gas. At a temperature of 298 Kelvin and a pressure of 1.03 atm, a volume of 1.53 L of hydrogen gas is collected. Assuming that the hydrochloric acid only reacts with the aluminum in the sample, what is the percent of aluminum in the alloy sample? (Solid aluminum reacts with hydrochloric acid to produce aluminum chlo-1. 36.6 percent

2. 50.4 percent

3. 10.5 percent

4. 72.0 percent

5. 21.7 percent

Option 5. 21.7 %

Explanation:

Let's think the reaction:

2Al (s) + 6HCl (l) → 3H₂ (g) + AlCl₃ (aq)

We have the data of hydrogen, we formed so let's apply the Ideal Gases Law equation to solve the amount of gas.

P. V = n. R. T

1.03 atm. 1.53L = n. 0.082 L. atm. 298K

(1.03 atm. 1.53L) / (0.082 L. atm. 298K) = n → 0.0645 moles of H₂

In the reaction, 3 moles of H₂ are produced by 2 moles of Al. Let's determine, how many moles of Al produced, the 0.0645 moles of H₂.

3 moles of H₂ were produced by 2 moles of Al

Then, 0.0645 moles of H₂ would be produced by (0.0645.2) / 3 = 0.043 moles.

If we convert the moles to mass, we can know the mass of Al in the alloy. (mol. molar mass)

0.043 m. 26.98 g/mol = 1.16 g

As the sample had a mass of 5.34 g, let's determine the % of Al.

(1.16 g / 5.34 g). 100 = 21.7%