Asked by kelly
A 1.500-g sample of a mixture containing only Cu2O and CuO was treated with hydrogen to prduce 1.252 g pure copper metal. Calculate the percentage composition (by mass) of the mixture.
Answers
Answered by
DrBob222
Let X = mass Cu2O
Let Y = mass Cu
===================
X + Y = 1.500
X(2Cu/Cu2O) + Y(Cu/CuO) = 1.252
Cu, Cu2O, and CuO mean molar masses.
Solve the two equations simultaneously for X and Y,
Then %Cu2O = (mass Cu2O/mass sample)*100
%Cu = (Mass Cu/mass sample)*100
Post your work if you get stuck.
Let Y = mass Cu
===================
X + Y = 1.500
X(2Cu/Cu2O) + Y(Cu/CuO) = 1.252
Cu, Cu2O, and CuO mean molar masses.
Solve the two equations simultaneously for X and Y,
Then %Cu2O = (mass Cu2O/mass sample)*100
%Cu = (Mass Cu/mass sample)*100
Post your work if you get stuck.
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