Exercise 10 | Chemical Calculations
15.0 g of copper are reacted with an excess of chlorine gas. If the percent yield is 87.5%, how many grams of copper(II) chloride are obtained?
Cu + Cl2 → CuCl2
From the chemical equation: 1 mole of CuCl2 is produced for each mole of Cu.
nCuCl2 = nCu
nCuCl2 = = =2.36 x 10-1 mol
mCuCl2 = nCuCl2 x MCuCl2 = 2.36 x 10-1 x 134.45 = 31.8 g
% yield = mobtained / mtheoretical x 100 = mobtained / mCuCl2 x 100
⇒ mobtained = 0.875 x 31.8 = 27.8 g