Answer to Question #248158 in General Chemistry for Hidra

Question

1.35g of aluminum powder and 8.0 g of copper (II) oxide were used in excess and heated vigorously in the laboratory to produce aluminum oxide and copper.

Calculate the mass of copper (ii) oxide not used in the reaction.

Answer 1

2Al + 3CuO → 3Cu + Al₂O₃

M(Al) = 26.98 g/mol

n(Al) $= \frac{1.35}{26.98} = 0.05 \;mol$

According to the reaction:

n(CuO) $= \frac{3}{2}n(Al) = \frac{3}{2} \times 0.05 = 0.075 \;mol$

M(CuO) = 79.54 g/mol

m(CuO) $= 0.075 \times 79.54 = 5.965 \;g$

Answer: 5.96 g