Transcript 10 - iannonechem.com

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
Ωg
Cu2O
(8.828 – Ω) g
CuO
Ω g Cu2O x 1mol x 2 mol Cu
x 63.5 g
143.0 g
1 mol Cu2O
1 mole
=
(8.828 – Ω) g CuO
= 7.051 - 0.7987 Ω g Cu
0.8881 Ω g +
0.0894 Ω
x 1mol x 1 mol Cu x
79.5 g
1 mol CuO
7.051
+ 7.051
-
0.7987 Ω g
63.5 g
1 mole
0.8881 Ω g Cu
= total grams Cu
= total grams Cu
Grams Cu
0.0894 Ω
+
7.051
0.0894 Ω
Ω
8.828
–
Ω
=
Grams Cu
=
7.214
=
0.163
=
mass Cu2O
=
1.823 g
=
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
Ωg
Ni2(SO4)3
Ω g NiSO4 x 1mol x 1 mol Ni
x 58.7 g
154.8 g
1 mol NiSO4
1 mole
(24.44 – Ω) g
=
0.3792 Ω g Ni
(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
0.0898 Ω
-
+ 7.072
0.2894 Ω g
= total grams Ni
= total grams Ni
Grams Ni
0.0898 Ω
+ 7.072
0.0898 A
Ω
24.44
–
Ω
=
Grams Ni
=
7.949
=
0.877
=
mass NiSO4
=
9.77 g
=
mass Ni2(SO4)3
=
14.67 g
A has 3 sig figs - molar masses!
Ni2(SO4)3 has 4 sig figs!
24.44
- 9. 7 7
14.67
Round to 2nd decimal