Question

Concerning the reaction: 2 HCl(aq) + Zn(s)  H2(g) + ZnCl2(aq) A piece of Zn with a mass of 3.2 g is placed in 233 mL of 0.19 M HCl (aq). What mass in grams of H2(g) is produced in the cases of: a) 100. % yield b) 63.0 % yield

Answer 1

Number of moles of Zn = 3.2g / 65.38g/mol = 0.0489 mole

molarity of HCl = number of moles of HCl / volume of solution in L

0.19 = number of moles of HCl / 0.233

number of moles of HCl = 0.0443 mole

from the balanced equation we can say that

2 mole of HCl requires 1 mole of Zn so

0.0443 mole of HCl will require 0.0222 mole of Zn

so HCl is the limiting reactant

2 mole of HCl produces 1 mole of H2 so

0.0443 mole of HCl will produce 0.0222 mole of H2

1 mole of H2 = 2.016 g

0.0222 mole of H2 = 0.0446 g

Therefore, theoretical yield of H2 = 0.0446 g

percent yield = (actual yield/theoretical yield)*100

100 = (actual yield / 0.0446)*100

actual yield of H2 = 0.0446g

Therefore, the mass of H2 produced when the reaction is 100% is 0.0446 g

percent yield = (actual yield/theoretical yield)*100

63 = (actual yield / 0.0446)*100

actual yield of H2 = 0.0446g

Therefore, the mass of H2 produced when the reaction is 63% is 0.0281 g