1. = 0.8881 Ω g Cu x 63.5 g 1 mole x 2 mol Cu 1 mol Cu 2 O x 1mol 143.0 g Ω g Cu 2 O = 7.051 - 0.7987 Ω g Cux 63.5 g 1 mole x 1 mol Cu 1 mol CuO x 1mol 79.5 g (8.828 – Ω) g CuO 0.8881 Ω g + 7.051 - 0.7987 Ω g= total grams Cu Chemistry 11 Challenge Question A mixture of Cu 2 O 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 Cu 2 O Cu 2 O Ω gCuO 0.0894 Ω + 7.051 = total grams Cu (8.828 – Ω) g

2. Grams Cu=Grams Cu 3 sig figs due to the molar masses! 8.828 – Ω = mass CuO = 7.0047 g Ω = mass Cu 2 O =1.823 g 0.0894 Ω = 0.163 0.0894 Ω + 7.051 = 7.214 % CuO = 7.0047 g x 100 % 8.828 g = 79.3 % Do not round until the end!

3. = 0.3792 Ω g Ni x 58.7 g 1 mole x 1 mol Ni 1 mol NiSO 4 x 1mol 154.8 g Ω g NiSO 4 = 7.072 - 0.2894 Ω g Nix 58.7 g 1 mole x 2 mol Ni 1 mol Ni 2 (SO 4 ) 3 x 1mol 405.7 g (24.44 – Ω) g Ni 2 (SO 4 ) 3 0.3792 Ω g + 7.072 - 0.2894 Ω g= total grams Ni Let Ω equal the mass of NiSO 4 NiSO 4 Ω g Ni 2 (SO 4 ) 3 0.0898 Ω + 7.072 = total grams Ni (24.44 – Ω) g 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?

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