= 0.8881 Ω g Cu x 63.5 g1 mole

x 2 mol Cu1 mol Cu2O

x 1mol143.0 g

Ω g Cu2O

= 7.051 - 0.7987 Ω g Cux 63.5 g1 mole

x 1 mol Cu1 mol CuO

x 1mol79.5 g

(8.828 – Ω) g CuO

0.8881 Ω g + 7.051 - 0.7987 Ω g = total grams Cu

Chemistry 11 Challenge QuestionA 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

Cu2O Ω g CuO

0.0894 Ω + 7.051 = total grams Cu

(8.828 – Ω) g

Grams Cu = Grams Cu

3 sig figs due to the molar masses!

8.828 – Ω = mass CuO = 7.0047 g

Ω = mass Cu2O = 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!

= 0.3792 Ω g Ni x 58.7 g1 mole

x 1 mol Ni1 mol NiSO4

x 1mol154.8 g

Ω g NiSO4

= 7.072 - 0.2894 Ω g Nix 58.7 g1 mole

x 2 mol Ni 1 mol Ni2(SO4)3

x 1mol 405.7 g

(24.44 – Ω) g Ni2(SO4)3

0.3792 Ω g + 7.072 - 0.2894 Ω g = total grams Ni

Let Ω equal the mass of NiSO4

NiSO4 Ω g Ni2(SO4)3

0.0898 Ω + 7.072 = total grams Ni

(24.44 – Ω) g

A container of nickel II sulphate has been accidentally contaminated withnickel 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?

Grams Ni = Grams Ni

A has 3 sig figs - molar masses!

Ω = mass NiSO4 = 9.77 g

0.0898 A = 0.877

0.0898 Ω + 7.072 = 7.949

24.44 – Ω = mass Ni2(SO4)3 = 14.67 g

2 4 . 4 4

1 4 . 6 7

- 9. 7 7 Ni2(SO4)3 has 4 sig figs!

Round to 2nd decimal