Chemistry Mixture Problems.. =( Please helpp mee?

Question

A mixture of magnesium and iron is treated with sulfuric acid to produce hydrogen gas and metal sulfates in solution according to the following equations:
2 Fe (s) + 3 H2SO4 (aq) --> Fe2(SO4)2 (aq) + 3 H2(g)
Mg (s) + H2SO4 (aq) --> MgSO4 (aq) + H2 (g)

When 4.413g of the mixture is treated with an excess of sulfuric acid, .2883g of hydrogen gas is collected. from this information, determine the percentage by mass of iron in the mixture.

I'd really appreciate it if you could explain to me what to do. =( I hate chem *tear*

Answer 1

Suppose there are x mol of Fe and y mol of Mg.
The total weight of the mixture will be 55.85 x + 24.31 y grams, so we have
55.85 x + 24.31 y = 4.413
Also, every mole of Fe produces 3/2 moles of H2, and every mole of Mg produces 1 mole of H2. So the number of moles of H2 produced is (3/2) x + y.

The mass of H2 collected was 0.2883 g, so
n(H2) = 0.2883 / (2*1.008) = 0.1430 mol.
So we have (3/2) x + y = 0.1430
or y = 0.1430 - (3/2) x.
Substitute this into the earlier equation
55.85 x + 24.31 y = 4.413
=> 55.85 x + 24.31(0.1430 - (3/2) x) = 4.413
=> 55.85 x + 3.476 - 36.46 x = 4.413
=> 19.39 x = 0.937
=> x = 0.0483
So there are 0.0483 mol of Fe in the sample.

Now m(Fe) = n(Fe) * 55.85 = 0.0483 (55.85) = 2.698 g.
So the percentage of Fe in the sample is
2.698 / 4.413 * 100 = 61.1%.

Answer 2

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RE:
Chemistry Mixture Problems.. =( Please helpp mee?
A mixture of magnesium and iron is treated with sulfuric acid to produce hydrogen gas and metal sulfates in solution according to the following equations:
2 Fe (s) + 3 H2SO4 (aq) --> Fe2(SO4)2 (aq) + 3 H2(g)
Mg (s) + H2SO4 (aq) --> MgSO4 (aq) + H2 (g)

When 4.413g of the mixture is...

Answer 3

I'll start you off - but I must say that I have never seen this type of question before in 28 years of teaching the subject. That is not to say that it's a bad question. On the contrary!

Let the moles of iron = x, and the moles of magnesium = y.

We can now set up two simultaneous equations:

56x + 24y = 4.413 and

1.5x + y = 0.2883/2

If you solve for x and y, and turn them back into mass ratios, you can get the percentage by mass of iron in the mixture.

Answer 4

First of all there is an error in the first equation. It should be Fe2(SO4)3 rather than Fe2(SO4)2, but that does not affect the answer.

You should not be discouraged by this type of questions; it is a non-typical one. If you understand the concept of moles, the rest follows logically.

Suppose the mass of Mg is x; then the mass of Fe is 4.413 - x. Covert these masses into moles.
Moles of Fe are = (4.413-x)/55.85
Moles of Mg = x/24.31

1 mol Fe gives off 3/2 mol Hydgrogen
1 mol Mg gives off 1 mol H2

So, H2 produced because of Fe = (4.413-x)/55.85 * 3/2 mol
H2 produced because of Mg = x/24.31 mol

Total moles of H2 = (4.413-x)/55.85 * 3/2 + x / 24.31---------[1]

Also, given total moles of H2 = 0.2883 /(2 * 1.008)------------[2]

Equate the two quantities [1] and [2]:

(4.413-x)/55.85 * 3/2 + x / 24.31 = 0.2883 /(2 * 1.008)

Solve for x. x = 1.689 g = mass of Mg.
mass of Fe = 4.413 - x = 2.724 g.

Get the percentages: % Fe = (2.724/4.413) * 100 = 61.73 5
% Mg = 38.27 %.