1.2 g of copper powder on heatinf gave 6.0 g of copper oxide. In another experiment, copper oxide contained 20% oxygen. Show that these results illustrate the law of constant proportion.

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Solution

2Cu + O2 â†’ 2CuO
2moles of CuO is given by = 1 mole of oxygen.
no. of moles of CuO in 6 g =6/79 moles
6/79 moles of CuO will give = (6/79)Ã—(1/2)=0.037 moles
Mass of 0.037 moles of Oxygen= 0.037Ã—32=1.184 g
Thus % of OXYGEN in CuO = (1.184/6)Ã—100=20 % OF OXYGEN

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Hint : In first case % of oygen in

C u O = 20.14

% .In second case % of oxygen in

C u O = 20.14

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1.2 g of copper powder on heatinf gave 6.0 g of copper oxide. In another experiment, copper oxide contained 20% oxygen. Show that these results illustrate the law of constant proportion.

2Cu + O2 â†’ 2CuO2moles of CuO is given by = 1 mole of oxygen.no. of moles of CuO in 6 g =6/79 moles6/79 moles of CuO will give = (6/79)Ã—(1/2)=0.037 molesMass of 0.037 moles of Oxygen= 0.037Ã—32=1.184 gThus % of OXYGEN in CuO = (1.184/6)Ã—100=20 % OF OXYGEN

2Cu + O2 â†’ 2CuO
2moles of CuO is given by = 1 mole of oxygen.
no. of moles of CuO in 6 g =6/79 moles
6/79 moles of CuO will give = (6/79)Ã—(1/2)=0.037 moles
Mass of 0.037 moles of Oxygen= 0.037Ã—32=1.184 g
Thus % of OXYGEN in CuO = (1.184/6)Ã—100=20 % OF OXYGEN