On heating in air, 0.12 g of metal gave 0.20 g of its oxide. The carbonate and nitrate of the metal were found to contain 28.5% and 16.2% of the metal respectively. Calculate by applying law of constnat proportions, the masses of oxide of the metal that will be obtained by heating 10 g each of the carbonate and the nitrate.

On heating in air, 0.12 g of metal gave 0.20 g of its oxide. The carbo

1.	On heating in air, 0.12 g of metal gave 0.20 g of its oxide. The carbonate and nitrate of the metal were found to contain 28.5% and 16.2% of the metal respectively. Calculate by applying law of constnat proportions, the masses of oxide of the metal that will be obtained by heating 10 g each of the carbonate and the nitrate.
Answer» Solution :Mass of oxygen in metal oxide =(0.20-0.12)=0.08g So, the ratio of masses of metal and oxygen in metal oxide 100gof metal CARBONATE contain metal =28.5g 10 G of metal carbonate contain metal `=(28.5)/(100)xx10=2.85g` If x g of oxygen combine with 2.85 g of metal, the ratio of masses of the metal and oxygen should be 3:2 as the LAW of constant proportions is true. `(2.85)/(x)=(3)/(2)` or `x=1.9g` Mass of metal oxide obtained from 10G metal carbonate =Mass of metal +Mass of oxygen `=(2.85+1.9)=4.75g` 100g of metal NITRATE contain metal =16.2g 10g of metal nitrate contain metal `=(16.2)/(100)xx10=1.62g` If y g oxygem combine with 1.62 g of metal, the ratio of masses of the metal and oxygen should be 3:2. `(1.62)/(y)=(3)/(2)` or `y=1.08g` Mass of metal oxide obtained from 10g metal nitrate =Mass of metal+Mass of oxygen. =(1.62+1.08)=2.70g