Copper forms two oxides, Cu2O and CuO - AQA - GCSE Chemistry - Question 8 - 2019 - Paper 1
Question 8
Copper forms two oxides, Cu2O and CuO.
A teacher investigated an oxide of copper.
Figure 10 shows the apparatus.
This is the method used.
1. Weigh empty tube A.
2.... show full transcript
Worked Solution & Example Answer:Copper forms two oxides, Cu2O and CuO - AQA - GCSE Chemistry - Question 8 - 2019 - Paper 1
Step 1
Suggest one reason why step 8 is needed.
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Step 8 is needed to ensure that all of the oxide of copper has reacted, guaranteeing that the experimental results accurately reflect the amount of copper produced.
Step 2
Explain why the excess hydrogen must be burned off.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The excess hydrogen must be burned off to prevent it from escaping into the air, as hydrogen is explosive. Burning off the excess ensures a safe and controlled environment during the experiment.
Step 3
Determine the mass of copper and the mass of water produced in this experiment.
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Using Table 5,
Mass of copper = 8.66 g
Mass of water = 2.45 g
Step 4
The teacher repeated the experiment with a different sample of the oxide of copper.
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To determine the correct equation for the reaction:
Using the data provided:
From the teacher's findings:
Mass of copper produced = 2.54 g
Mass of water produced = 0.72 g
Using the molar masses:
For Equation 1 (Cu2O + H2 → 2 Cu + H2O):
Moles of Cu = 2.54 g / 63.5 g/mol = 0.04 mol (producing 0.04 mol of H2O, which is 0.04 x 18 = 0.72 g of water)
Matches the observed results, so Equation 1 is correct.
For Equation 2 (CuO + H2 → Cu + H2O):
Moles of H2O = 0.72 g / 18 g/mol = 0.04 mol
This gives 0.04 mol Cu, which would produce 2 times that amount in H2O, hence confirming Equation 2 correctly balances as well.
Thus, both equations can potentially fit depending on the amount of the oxide used.
