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
The nuclear fuel used in a nuclear reactor needs to have a higher concentration of the U 235 isotope than that which exists in natural uranium ore. U235 when concentrated (or "enriched") is fissionable in light-water reactors (the most common reactor design in the USA).
