What weight of zinc would be required to produce
enough hydrogen to reduce completely 8.5 g of
copper oxide to copper ?
Answers
Solution:-
Reaction of zinc to produce hydrogen-
Zn+2HCl⟶ZnCl
2
+H
2
Reaction of hydrogen to reduce copper oxide into copper-
CuO+H
2
⟶H
2
O+Cu
From the above reactions-
1 moles of Zn produces 1 mole of hydrogen gas and 1 mole of hydrogen required to reduce 1 mole of CuO.
Mole of CuO reduced = mole of Zn required
Therefore,
No. of moles of CuO reduced =
molar mass of CuO
moles of CuO
Given weight of CuO=8.5g
Molecular weight of CuO=79.5g
Therefore,
No. of moles of CuO reduced =
79.5
8.5
=0.107 mol
Therefore,
No. of moles of Zn required =0.107 mol
Now,
Molecular weight of Zn=65.4g
∴ Weight of Zn required =0.107×65.4=6.99g
Hence 6.99g of Zn will be required.
Answer:
ANSWER
Solution:-
Reaction of zinc to produce hydrogen-
Zn+2HCl⟶ZnCl
2
+H
2
Reaction of hydrogen to reduce copper oxide into copper-
CuO+H
2
⟶H
2
O+Cu
From the above reactions-
1 moles of Zn produces 1 mole of hydrogen gas and 1 mole of hydrogen required to reduce 1 mole of CuO.
Mole of CuO reduced = mole of Zn required
Therefore,
No. of moles of CuO reduced =
molar mass of CuO
moles of CuO
Given weight of CuO=8.5g
Molecular weight of CuO=79.5g
Therefore,
No. of moles of CuO reduced =
79.5
8.5
=0.107 mol
Therefore,
No. of moles of Zn required =0.107 mol
Now,
Molecular weight of Zn=65.4g
∴ Weight of Zn required =0.107×65.4=6.99g
Hence 6.99g of Zn will be required
Explanation:
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