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2.75 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was 2.196 g. Another experiment, 2.358 g of copper was dissolved in nitric acid and the resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed was 2.952 g. Show that these results illustrate the law of constant composition.

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Hint:According to the law of constant composition the compound contains the same number of elements and the same mass ratio. In both the experiments determine the mass of oxygen formed by the reduction of copper oxide into copper and oxygen.

Complete step by step answer:
The mass of cupric oxide in experiment 1 is 2.75 g
The mass of copper remained in experiment 1 is 2.196 g.
The mass of copper in experiment 2 is 2.358 g.
The mass of cupric oxide is 2.952 g.
Experiment 1-
The reduction reaction of cupric oxide is shown below.
$CuO\xrightarrow{\Delta }Cu + \dfrac{1}{2}{O_2}$
The one mole of cupric oxide on heating gives one mole of copper and half mole of oxygen.
In this experiment, 2.75 g of cupric oxide is reduced by heating and the remaining mass of copper is 2.196 g.
Let assume the mass of oxygen as W.
The mass of oxygen is calculated as shown below.
$\Rightarrow$2.75 g = 2.196 + W
$\Rightarrow$W = 2.75 g – 2.196
$\Rightarrow$W = 0.554 g.
Experiment 2-
Copper is dissolved in nitric acid and the product copper nitrate is converted to cupric oxide.
The reaction is shown below.
$Cu + HN{O_3} \to Cu{(N{O_3})_2} \to CuO$
2.358 g of Cu is dissolved in water. The copper oxide formed by ignition of copper nitrate is 2.952 g.
The dissociation reaction of copper oxide is shown below.
$CuO \to Cu + \dfrac{1}{2}{O_2}$
Let assume the mass of oxygen as W.
The mass of oxygen is calculated as shown below.
$\Rightarrow$2.952 g = 2.358 + W
$\Rightarrow$W = 2.952 g – 2.358
$\Rightarrow$W = 0.594 g.
The law of constant composition states that the compound contains the same number of elements in the same mass ratio.
Here the mass of oxygen formed in both the experiments is almost the same. Therefore, it follows the law of constant composition.
Therefore, the correct option is A.

Note:
The answer can also be determined by calculating the mass percent of the compound by dividing the mass of copper by mass of copper oxide multiplied by 100.