Answer:
88%.
Explanation:
The percentage yield of lead sulfate in this experiment is 88% if 2.53 is divided by 2.85 and multiply by 100. The percentage yield can be calculated when the experimental yield is divided by theoretical yield and then multiply by 100. The percentage yield tells us about the actual yield that is gained in the end of experiment which is lower than theoretical yield.
Answer: 1 mol of
will be produced from this reaction.
Explanation: Reaction follows,

As seen from the balanced chemical equation above, we get
For every 3 moles of Aluminium and 3 moles of
, 1 mole of
is formed.
For every 3 moles of Aluminium and 3 moles of
, 1 mole of
is formed.
For every 3 moles of Aluminium and 3 moles of
, 3 moles of
is formed.
For every 3 moles of Aluminium and 3 moles of
, 6 moles of
is formed.
a)
A: Copper
B: CuO
C: 
D: $\mathrm{CuCO_3}$
E: $\mathrm{CO_2}$
F: $\mathrm{Cu(NO_3)_2}$
b)
$\mathrm{CuO+ H_2SO_4}\rightarrow \mathrm{CuSO_4 + H_2O}$
c)
$\mathrm{CuCO_3+ 2HNO_3}\rightarrow \mathrm{Cu(NO_3)_2+ CO_2+ H_2O}$