Calculating Mass Percent of Aluminum in a Sample Using Ideal Gas Law

[RaamGeneral]

Hello, sorry to bother again.

A sample containing Al reacts with an aqueous solution of 5.05 g HCl producing 5.70 L of H₂ at 20°C and 742 mmHg.
Find the mass percent of Al in the sample assuming hydrogen behaves like an ideal gas.I can't again understand this exercise, because I've never done any like this before.
The only reaction I know involving aluminium and HCl is:
\mathrm{2Al+6HCl \to 2AlCl_3 + 3H_2}
but it doesn't seem useful.

I calculated 0.139 mol of Al and 0.231 mol of H₂ using the ideal gas law.

I can't see any clue about my sample though.

 

[Borek]

Your number of moles of hydrogen is correct, yet you managed to miscalculate the amount of Al.

As worded question can't be solved, and it is self contradicting - you can't produce 0.231 moles of hydrogen using 5.05 grams of HCl. My bet is 5.05 g is a mass of the Al sample. Would it help?

 

[RaamGeneral]

My bad, what a mess: I meant 0.139 mol of HCl. But let's forget about this because your bet is correct: I saw again the text of the exercise and I didn't noticed a comma, it was 5.05 g of the sample containing Al (not only it).

Nevertheless I'm not able to understand the exercise yet: HCl reacts with this sample producing H₂. The reaction I wrote earlier doesn't seem useful here.
I don't know how much HCl reacts because I don't know the substances involved in the reaction and so the stoichiometric coefficients.

It will be wrong, but my interpretation of the problem is this:

\mathrm{HCl+X...Al_n \to Y+H_2}Thank you for your answer.

 

[Borek]

The reaction you wrote earlier is exactly the one needed to solve the problem.

Doesn't matter how much HCl reacted, only thing that matters is that all Al was dissolved. That means amount of hydrogen produced depends on the amount of Al present initially.

 

[RaamGeneral]

From the stoichiometry and considering 0.231 mol of H₂ are formed, the moles of Al reacted are 0.154, that is 4.16 g (82.4% of the sample).

Is it right?

 

[Borek]

Yes, that's it.