`6g` of a hydrocarbon on combustion in excess of oxygen produces `17.6g` of `CO_(2)` and `10.8g` of `H_(2)O`. The data illustrates the law of `:`

Options

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

Answer

1

Explanation

`{:(C_(X)H_(Y)+,,O_(2),rarr,CO_(2),+,H_(2)O),(6g,,excess,,,,):}`
mass of Hydrocarbon `=6g`
mass of carbon `=(12)/(44)xx` mass of `CO_(2)=(12)/(44)xx17.6=4.8g`
mass of Hydrogen `=(2)/(18)xx` mass of `H_(2)O=(2)/(18)xx10.8=1.2g`
Total mass of carbon & Hydrogen `=4.8+1.2=6g`
Mass of Hydrogen `=` Total mass of `(C+H)=6g`.
Law of conservation of mass.