What Is the Percent Composition of Magnesium and Aluminum in the Alloy?

[johndoe]

A metal alloy consists of magnesium and aluminum.When a 1.00g sample of the alloy is reacted with excess HCl, 0.0989g of hydrogen gas forms. Calculate the percent composition of the alloy.

Ok I got the following balanced equations :

Mg + 2HCl -> MgCl2 + H2
2Al + 6Hcl -> 2AlCl3 + 3H2

Can I say 1 mole of Mg reacts with 1 Mole of H2, with H2 being 1/4 of 0.0989g and
2 moles of Al reacts with 3 moles of H2, with H2 being 3/4 of 0.0989g?

 

[Borek]

No, you can't. That will happen only if you will know you have mixture of 1 mole of Mg and 2 moles of Al, you don't know neithe rhow many moles of both nor in waht ratio you have.

Write two equations - one for sum of masses of metals, one for sum of numbers of moles of metals. The latter will use moles of hydrogen produced and stoichiometry of the reactions provided.

 

[Blueshift5]

Yes, your approach is correct. The balanced equations show that for every mole of Mg reacted, 1 mole of H2 is produced, and for every 2 moles of Al reacted, 3 moles of H2 are produced. Therefore, we can use the molar mass of H2 (2.016 g/mol) to calculate the moles of Mg and Al present in the alloy.

Using the given mass of hydrogen gas (0.0989g), we can calculate the moles of H2 produced:

0.0989g H2 x (1 mole H2/2.016g H2) = 0.0491 moles H2

Since H2 is 1/4 of the alloy, we can multiply the moles of H2 by 4 to get the total moles of the alloy:

0.0491 moles H2 x 4 = 0.1964 moles alloy

Now we can use the molar mass of each element to calculate the mass of Mg and Al in the alloy:

Mg: 0.1964 moles x 24.31 g/mol = 4.77g
Al: 0.1964 moles x 26.98 g/mol = 5.29g

Therefore, the alloy is composed of 4.77g of Mg and 5.29g of Al, making the percent composition:

%Mg = (4.77g/10.06g) x 100% = 47.4%
%Al = (5.29g/10.06g) x 100% = 52.6%

In conclusion, the alloy is composed of approximately 47.4% magnesium and 52.6% aluminum.