Chemistry question!
would really appreciate!

A mixture of Cu2 and CuO of mass 8.828g is reduced to copper metal with hydrogen:
Cu2O + H2 --> 2Cu + H2O
CuO + H2 --> Cu + H2O

If the mass of pure copper isolated was 7.214g, determine the percent by mass of CuO in the original sample

Let x = grams of CuO in the original sample.
y = grams of Cu2O in the original sample.

Eq. #1 x + y = 8.828 grams

Molar mass of CuO = 63.5 + 16 = 79.5 grams
Moles of CuO = x ÷ 79.5

Molar mass of Cu2O = 63.546 + 32 = 95.5 grams
Moles of Cu2O = y ÷ 95.5

According to the 2nd balanced equation, CuO + H2 --> Cu + H2O ,
1 mole of CuO produces 1 mole of Cu.
So, x ÷ 79.5 moles of CuO will produce x ÷ 79.5 moles of Cu

According to the 1st balanced equation, Cu2O + H2 --> 2Cu + H2O,
1 mole of Cu2O produces 2 moles of Cu
So, (y ÷ 95.5) moles of Cu2O will produce 2 * (y ÷ 95.5) moles of Cu

Since, the mass of pure copper isolated was 7.214 grams
Moles of Cu = (7.214 ÷ 63.5)

Moles of Cu from Cu2O + moles of Cu from CuO = total moles of Cu!!

2 * (y ÷ 95.5) + (x ÷ 79.5) = (7.214 ÷ 63.5)
Multiply by both sides by 95.5 * 79.5 * 63.5 to get rid of denominators

(2 * 79.5 * 63.5) y + (95.5 * 63.5) x = (7.214 * 95.5 * 79.5)

10,096.5 y + 6,064.25 x = 36,418.0755
Divide both sides by 6,064.25
x + 1.665 y = 6

Eq.#2 x = 6 – 1.665 y
Eq. #1 x + y = 8.828
x = 8.828 – y

8.828 – y = 6 – 1.665 y
0.665 y = 2.828
y = 4.25 grams of Cu2O
x = 8.828 – 4.25 = 4.58 grams of CuO

% CuO = (4.58 ÷ 8.828) * 100 = 51.88% CuO

Noble gases is far right