Answer to Question #62669 in General Chemistry for Jenna

Question

How do I calculate this since the H2 is coming from two separate sources.

A 4.439-g sample of a mixture of Fe and Al is treated with excess HCl(aq). If 0.1915 moles of H2 are obtained, then what is the percentage by mass of Fe in the original mixture?


Fe(s) + 2 HCl(aq) --> FeCl2(aq) + H2(g)
2 Al(s) + 6 HCl(aq) --> 2 AlCl3(aq) + 3 H2(g)

Answer 1

From the reaction equations we can see, that 1 mole of Fe produces 1 mole of hydrogen, and 1 mole of Al produces 3/2 moles of hydrogene. 1 mole of Fe has a mass of 55.849 g; one mole of Al - 26.092 g.

Assume that amount of moles of Fe in the mixture is X, and the amount of moles of Al is Y.
Then the total mass of mixture is 55.849 * X + 26.092 * Y and it is 4.439 g.
And the amount of hydrogene obtained is X + 3/2 * Y and it is 0.1915 moles.

Solve these two equations:
55.849 * X + 26.092 * Y = 4.439 (1)
X + 1.5Y = 0.1915 (2)
Express Y from (2):
Y = (0.1915 – X) / 1.5.
Put Y into (1) and do the calculations:
55.849 * X + 26.092 * ((0.1915 – X) / 1.5) = 4.439;
55.849X + 3.331 – 17.395X = 4.439;
38.454X = 1.108;
X = 0.029.

So the mixture contains 0.029 moles of Fe which is 0.029 * 55.849 = 1.620 g.
The percentage by mass of Fe in the original mixture is (mass of Fe / mass of mixture) * 100% = (1.620 g / 4.439 g) * 100% = 36.5%

Answer:
The percentage by mass of Fe in the original mixture is 36.5%