Stoichiometry problem with excess reactant & 95% completion?

Question

I need a little bit of help for the following problem...

We wish the analyze the percentage of copper and zinc in a post 1982 penny. To do so, we weigh out a penny and find it to be 3.108 g (+/- 0.001 g).

a. Which of the two metals will react with hydrochloric acid (HCL)? Write the balanced reaction.

b. How many mL of 6.53 M HCL will it take to react, assuming the entire penny to be zinc (even though it's not)?

c. If excess HCL is used, and the reaction goes to 95.0% completion, we find that 887 mL of hydrogen gas is collected at STP. How many grams of each metal is contained in the penny, and what is the percentage of each metal in the penny?

..I've done most of it, but I'm a little stuck on part c. If anyone could review my work and give me a nudge in the right direction, it would be much appreciated.

a.) 2 Zn (s) + 2 HCL (aq) ----> H2 (g) + 2 ZnCl (aq)

b.) M= moles / L

L= mol HCl / M

3.108 g Zn (1 mol Zn / 65.38 g Zn) (2 mol HCl / 2 mol Zn) = .04754 mol HCl

L = .04754 mol HCl / 6.53 M HCl = .00728 L HCl = 7.28 mL HCl

c.) 887 mL = .887 L

.887 L H2 (1 mol H2 / 22.4 L) (2 mol Zn / 1 mol H2) (65.38 g Zn / 1 mol Zn) = 5.18 g Zn.

....Obviously this can't be right because that's more than the mass of the penny. It's possible that my teacher could have made a mistake and put in numbers that don't work, but my main question is, how does the 95% completion fit in to the problem? I haven't done a problem like that before.

any help would be appreciated. :) thanks!

Answer 1

You are on the right track, all you working is good. Your problem was the initial equation

Zn + 2HCl -------> H2(g) + ZnCl2

So Zn and HCl react 1:2 ratio

And 1 mole H2 forms from 1 mol Zn

Also, don't forget that you only got 95% reaction. So the amount of Zn you work out only be 95% of what is actually there.

From what I can see the rest of your working is good. So have another go with the correct equation.

Answer 2

it is your equation, Zn forms a Zn+2  ion which makes ZnCl2, & doubles  answer "b" 7 cuts answer "c" in half:

a.) 1 Zn (s) + 2 HCL (aq) ----> H2 (g) +  ZnCl2 (aq)

==========================

b.) M= moles / L

L= mol HCl / M

3.108 g Zn (1 mol Zn / 65.38 g Zn) (2 mol HCl /  mol Zn) = 0.09508 mol HCl

L = 0.09508 mol HCl / 6.53 M HCl = 0.01456 L HCl = 14.56 mL HCl

============================

c.) 887 mL = .887 L

.887 L H2 (1 mol H2 / 22.4 L) ( 1 mol Zn / 1 mol H2) (65.38 g Zn / 1 mol Zn) = 2.59 g Zn.

95% yield:

2.59 g Zn @ 100/95 = 2.726 grams of Zn

find grams of copper:

3.108 g - 2.726 g Zn = 0.382 grams of copper

%:

0.382 / 3.108 times 100 = 12.3 % Cu

your answers are 0.38 grams of Cu  & 2.73 grams Zn

your answers are 12.3% Cu  & 87.7% Zn