Chemistry 11 Challenge Question

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Chemistry 11 Challenge Question A mixture of Cu2O and CuO of mass 8.828 g is reduced to copper metal with hydrogen. If the mass of pure copper isolated was 7.214 g, determine the percent (by mass) of CuO in the original sample. Let Ω equal the mass of Cu2O (8.828 – Ω) g Cu2O Ω g CuO Ω g Cu2O x 1mol x 2 mol Cu x 63.5 g = 0.8881 Ω g Cu 143.0 g 1 mol Cu2O 1 mole (8.828 – Ω) g CuO x 1mol x 1 mol Cu x 63.5 g = 7.051 - 0.7987 Ω g Cu 79.5 g 1 mol CuO 1 mole 0.8881 Ω g + 7.051 - 0.7987 Ω g = total grams Cu 0.0894 Ω + 7.051 = total grams Cu

Grams Cu = Grams Cu 0.0894 Ω + 7.051 = 7.214 0.0894 Ω = 0.163 Ω = mass Cu2O = 1.823 g 8.828 – Ω = mass CuO = 7.0047 g Do not round until the end! % CuO = 7.0047 g x 100 % = 79.3 % 8.828 g 3 sig figs due to the molar masses!

A container of nickel II sulphate has been accidentally contaminated with nickel III sulphate. The total mass of both sulphates was 24.44 g. Through a single replacement reaction with Zn, the nickel was extracted from both sulphates and was found to have a mass of 7.949 g. What was the original masses of the nickel II sulphate and nickel III sulphate before they were mixed? Let Ω equal the mass of NiSO4 NiSO4 Ω g Ni2(SO4)3 (24.44 – Ω) g Ω g NiSO4 x 1mol x 1 mol Ni x 58.7 g = 0.3792 Ω g Ni 154.8 g 1 mol NiSO4 1 mole (24.44 – Ω) g Ni2(SO4)3 x 1mol x 2 mol Ni x 58.7 g = 7.072 - 0.2894 Ω g Ni 405.7 g 1 mol Ni2(SO4)3 1 mole 0.3792 Ω g + 7.072 - 0.2894 Ω g = total grams Ni 0.0898 Ω + 7.072 = total grams Ni

2 4 . 4 4 - 9. 7 7 1 4 . 6 7 Round to 2nd decimal Grams Ni = Grams Ni 0.0898 Ω + 7.072 = 7.949 0.0898 A = 0.877 Ω = mass NiSO4 = 9.77 g 24.44 – Ω = mass Ni2(SO4)3 = 14.67 g 2 4 . 4 4 - 9. 7 7 Round to 2nd decimal A has 3 sig figs - molar masses! 1 4 . 6 7 Ni2(SO4)3 has 4 sig figs!