Question

12.0 g of copper is concerted to copper (II) oxide by heating in oxygen.

Write a balanced equation for this reaction.
What mass of oxygen is required to convert the 12.0 g of copper into copper (II) oxide?
What mass of copper (II) oxide will theoretically be produced?
What is the percent yield of copper (II) oxide if 12.8 g of copper (II) oxide is actually produced?

Answer 1

Basic information:

Atomic mass of Copper (Cu) = 63.5 g/mol

Molecular mass of copper (II) oxide : 79.5 g/mol

Step 1: Balanced equation is

2Cu + O_{​​​​​​2} -------------> 2CuO

Step 2: Calculation of mass of oxygen.

According to the Balanced equation, 2 X 63.5 g/mol of copper reacts with 32 g/mol of oxygen, then the mass of oxygen required to react with 12.0 g of copper is...

Mass of oxygen = 12 g X 32 g/mol / 2 X 63.5 g/mol

Mass of oxygen = 3.023 g.

Step 3: Calculation of the mass of copper (II) oxide formed

2 X 63.5 g/mol of copper produces, 2 X 79.5 g/mol of CuO, then

12.0 g of Cu can produce, the mass of CuO

Mass of CuO formed = 12 g X 2 X 79.5 g/mol / 2 X 63.5 g/mol

Mass of CuO formed= 15.02 g.

Step 4: Calculation of percent yield of CuO.

Mass of CuO to be produced = 15.02 g

Mass of CuO actually produced = 12.8 g

% yield of CuO = 12.8 g X 100 / 15.02 g

% yield of CuO = 85.219 %.