8 Copper forms two oxides, Cu2O and CuO A teacher investigated an oxide of copper.
Figure 10 shows the apparatus
Figure 10
This is the method used.
Weigh empty tube A.
Add some of the oxide of copper to tube A.
Weigh tube A and the oxide of copper.
Weigh tube B and drying agent.
Pass hydrogen through the apparatus and light the flame at the end.
Heat tube A for 2 minutes.
Reweigh tube A and contents.
Repeat steps 5 to 7 until the mass no longer changes.
Reweigh tube B and contents.
Repeat steps 1 to 9 with different masses of the oxide of copper.
8-1 Suggest one reason why step 8 is needed.
ANSWER
To make sure all of the oxide of copper has reacted and also to make sure that all water (produced) is removed.
8-2 Explain why the excess hydrogen must be burned off.
[2 marks]
ANSWER
As hydrogen is explosive so it needs to be prevented from escaping into the air.
Figure 10 is repeated here
Figure 10
Table 5 shows the teacher’s results.
Table 5
When an oxide of copper is heated in a stream of hydrogen, the word equation for the reaction is:
Copper oxide + hydrogen → copper + water
8-3 Determine the mass of copper and the mass of water produced in this experiment. Use Table 5.
[2 marks]
ANSWER
Mass of water
8-4 The teacher repeated the experiment with a different sample of the oxide of copper. The teacher found that the oxide of copper produced 2.54 g of copper and 0.72 g of water.
Two possible equations for the reaction are:
Equation 1: Cu2O + H2 → 2 Cu + H2O
Equation 2: CuO + H2 → Cu + H2O
Determine which is the correct equation for the reaction in the teacher’s experiment.
Relative atomic masses (Ar): H = 1 O = 16 Cu = 63.5
ANSWER
Step 1: First calculate the moles of copper and water
Moles of Copper
Step 2: Identify the smallest number of moles from the calculated mole
Both have the same moles, therefore we will take 0.04
Step 3: Divide calculate moles of Copper and water by 0.04 mol
For Copper,
For water
Therefore, the ratio of Copper and water = 1:1
Hence, Equation 2 meets the requirement.
