Chemistry, asked by pranshauhanu, 10 months ago

verify the law of constant proportion for an experiment in which 2.06 g of copper forms 2.576 g of copper oxide and in another experiment 7.344 g of copper oxide reduces to form 5.876 g of copper

Answers

Answered by abbycookie10

Explanation:

Part One:

2Cu + O₂ ----> 2CuO

Mole ratio of copper : copper oxide

2 : 2

∴ 1 : 1

1 mole of Cu = 64g

1 mole of O = 16g

1 mole of Cu = 64g

∴ Number of moles in 2.06g of Cu = $\frac{2.06}{64}$

= 0.032 mol

1 mole of CuO = 80g

∴ Number of moles in 2.576g of CuO = $\frac{2.576}{80}$

= 0.032 mol

This shows that the law of constant proportion is being upheld.

Part Two:

1 mole of Cu = 64g

∴ Number of moles in 5.876g of Cu = $\frac{5.876}{64}$

= 0.0918 mol

1 mole of CuO = 80g

∴ Number of moles in 7.344g of CuO = $\frac{7.344}{80}$

= 0.0918 mol

This also shows that the law of constant proportion is being upheld.

Hope this helps! :)

- Abby <3

Answered by SHRIYANDE

Answer:

Explanation:

Part One:

2Cu + O₂ ----> 2CuO

Mole ratio of copper : copper oxide

2 : 2

∴ 1 : 1

1 mole of Cu = 64g

1 mole of O = 16g

1 mole of Cu = 64g

∴ Number of moles in 2.06g of Cu =

= 0.032 mol

1 mole of CuO = 80g

∴ Number of moles in 2.576g of CuO =

= 0.032 mol

This shows that the law of constant proportion is being upheld.

Part Two:

1 mole of Cu = 64g

∴ Number of moles in 5.876g of Cu = 0.0918 mol

1 mole of CuO = 80g

∴ Number of moles in 7.344g of CuO = 0.0918 mol

This also shows that the law of constant proportion is being upheld.

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