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.

Explanation

Mass of oxygen in metal oxide
=(0.20-0.12)=0.08g
So, the ratio of masses of metal and oxygen in metal oxide
100g of 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