Calculating percentage of Zn in a coin

Question

Question:

An old coin found in an ancient temple is composed of zinc coated with copper. In an experiment to find the percent zinc in the coin, a student determined the weight of the coin to be 3.0 g. Then the student made several scratches in the copper coating (to expose the underlying zinc) and put the scratched coin in hydrochloric acid, where the following reaction occurred between the zinc and HCl (copper remained undissolved):

Zn(s) + 2 HCl (aq) → H2(g) + ZnCl2 (aq)

The student collected the hydrogen produced over water at 27 °C. The collected gas occupied a volume of 1 L at a total pressure of 1.02 bar. The percent zinc in the coin, the student found would be. (Assume that all the Zn in the coin dissolves.)

( a) 67.7 %

(b) 98.2 %

(c) 96.7 %

(d) 88.3 %

(e) 25.0 %

My attempt:

P = 1.02 x 10^5 Pa

V = 1 x 10^-3 m^3

T = 27 + 273 = 300K

Use PV=nRT,

(1.02 x 10^5)(1 x 10^-3)=n x .31 x 300

n = 0.04091 moles of H2

moles of Zn = 0.04091

Mass of Zn = 0.04091 x 65.39 = 2.67 g

Percentage of Zn = (2.67/3.0) x 100 = 89.2%

The mark scheme answer is D) 88.3% which is the closest answer to the value I got, but why don't I get that answer? Is it an error in my calculation, the mcq answers or something else?

Answer 1

If the original data gives an original weight of $3.0$ g, it means that the real weight is situated between $2.95$ g and $3.05$ g. It also means that the precision on this weight is $±0.05$ g, which correspond to ± $0.05/3.0 = 1.7$%. The final result is also known with a precision of ±$1.7$%. And ±$1.7$% of $88.3$ is ±$1.47$ Take care of the following calculations. Don't confuse percent and percent of percent. The final value is something between $88,3 - 1.47 = 86.8$% and $88.3 + 1.47 = 89.7$%. As a consequence, your result ($89.2$%) is within these limits. It is valid.