Question

1.5 g of hydrocarbon on combustion in excess of oxygen produces 4.4 g of $C{O}_2$ and 2.7 g of $H_{2}O$, then the data illustrates:

• A. Law of conservation of mass
• B. Law of multiple proportions
• C. Law of constant composition
• D. Law of reciprocal proportions

Answer 1

The correct option is A Law of conservation of mass

Explanation:

Mass of hydrocarbon $C_x{H}_y=1.5g$

$C$ changes to $C{O}_2$ and $H$ changes to $H_{2}O$

$C$ in 4.4g $C{O}_2$ is $1.2g$ (in $44g$ --- $C{O}_2$ $12g$ of Carbon is present, so in $4.4g$ $1.2g$ carbon is present)

Hydrogen in $2.7g$ $H_{2}O=\frac{2\times 2.7}{18}=0.3g$

Mass of $C+H=1.2+0.3=1.5g$

Which is equal to the mass of hydrocarbon taken.

So this is law of conservation of mass, because here mass is conserved.

Hence option A is correct.