[Araconan]

Hi!

I am currently attempting the following problem:

Zinc and magnesium metal each react with hydrochloric acid according to the following equations:

Zn(s) + 2HCl(aq) -----> ZnCl2(aq) +  H2(g)
Mg(s) + 2HCl(aq) -----> MgCl2(aq) + H2(g)

A 10.00g mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with 156 mL of 3.00 M silver nitrate to produce the maximum possible amount of silver chloride.

Determine the percent magnesium by mass in the original mixture.

My attempt at the problem:

1) Since the original compound is a mixture of Zn and Mg, both of the above reactions will occur at once, when the compound is placed into the HCl. Therefore, I add the two chemical reactions together, as one big reaction between the mixture, and the HCl. After balancing:

Zn + Mg + 4HCl -----> ZnCl2 + MgCl2 + 2H2(g)                                    (1)

The hydrogen gas escapes, so the two products of Zinc Chloride and Magnesium Chloride will then react with Silver Nitrate. Again, I combine both of the reactions between ZnCl2 and AgNO3 and MgCl2 and AgNO3 into one equation. After balancing:

ZnCl2 + MgCl2 + 4AgNO3 -----> 4AgCl + Zn(NO3)2 + Mg(NO3)2               (2)

Converting 156 mL of 3.00M AgNO3 into moles = 0.468 moles of AgNO3. Referring back to equation 2, it would mean that there are 0.468/4 moles of ZnCl2 and MgCl2. Then referring to equation 1, it would mean that there was originally 0.468/4 moles of Zn and 0.468 moles of Mg. (Based upon mole ratios)

But if you were to covert this amount of moles of Mg and Zn into actual grams, you would get 2.84427g (Mg) and 7.64946g (Zn), which add up to 10.49373g, which is more than the 10g that was specified. Furthermore, this would calculate a magnesium mass percentage of 28.4427%, while the answer is 31.3%.

So I'm wondering, what am I doing wrong?

Thank you in advance!

[UG]

When you added the two equations together, you are assuming there are equal amounts of Zn and Mg, which is not the case. These kind of problems can usually be solved with two simultaneous equations, the first equation should be pretty easy to deduce, and that is:
m(Zn) + m(Mg) = 10 g               [1]
The second equation is a bit more tricky and you need the information given about silver nitrate to deduce it. Try and see if you can get the second equation in terms of the two variables in [1]. Think about moles and molar masses.

[Araconan]

Ahh I see where I went wrong now.

I followed your advice, and basically set up an equation for the number of chloride moles, since I could determine the amount of Chloride moles based upon calculating the amount of moles of Zn and Mg, and multiplying it by two, and since all of the Chloride ions end up in the Silver Chloride precipitate, I just converted the amount of Silver Chloride into moles. So,

2*(X/65.38) + 2*[(10-x)/24.31] = 0.468

Solved for x, and got the answer.

Thank you very much for your *delete me*

[Araconan]

Ahh I see where I went wrong now.

I followed your advice, and basically set up an equation for the number of chloride moles, since I could determine the amount of Chloride moles based upon calculating the amount of moles of Zn and Mg, and multiplying it by two, and since all of the Chloride ions end up in the Silver Chloride precipitate, I just converted the amount of Silver Chloride into moles. So,

2*(X/65.38) + 2*[(10-x)/24.31] = 0.468

Solved for x, and got the answer.

Thank you very much for your *delete me*

Hm, I didn't know that typing "Thankyou very much for your help" with a space in between 'thank' and 'you' would automatically be changed into "Thankyou very much for your *delete me*" O.o

Anyways, I meant to say thanks!

[Borek]

It is "help" that was eaten.