A 0.50 g mixture of Cu and CuO was treated with hydrogen at elevated temperature. CuO was

Question

reduced to Cu .

(b) Before the experiment, calculate the minimum amount (in mole) of hydrogen required
to ensure the reduction is completed. (3 marks)
(c) After passing with in excess amount of hydrogen gas, 0.42 g of solid remained.
Calculate the percentage of CuO in the mixture.

Answer 1

I wonder what part a to the question was. You can't do part b without doing part c first.

c.
(1) g Cu + g CuO = 0.5g
(2) g Cu + g Cu + g O = 0.5 g
After heating the O is gone so subtract (2) from (1) to get
g O = 0.5-0.42 = 0.08

The reduction equation is
2H2 + O2 ==> 2H2O
mols O2 = 0.08/32 = 0.0025
mols H2 = 2*0.0025 = 0.005

For percent CuO, you need to know how much CuO was in the original 0.5 g sample.
CuO + H2 ==> Cu + H2O
You know 0.005 mols H2 required which means 0.005 molsCuO initially. Convert to grams CuO, then to % CuO in the original sample.

Answer 2

Part (A) is just balance the equation that I had done it.

Answer 3

the answer is what you had thought