`3 g` of ethane `C_(2)H_(6)` on complete combustion gave `8.8 g` of `CO_(2)` and `5.4 g` of water. Show that the results are in accordance with the law of conservation of mass.

Explanation

`underset({:(12 xx 2 + 6),(= 30 g):})(C_(2) H_(6)) + underset(underset(= 122g)(32 xx (7)/(2)))((7)/(2) O_(2)) rarr underset({:(44 xx 2),(= 88 g):})(2CO_(2)) + underset({:(18 xx 3),(= 54 g):})(3H_(2) O)`
a. `30 g of C_(2) H_(6) = 88 g of CO_(2)`
`3 g of C_(2) H_(6) = 8.8 g of CO_(2)`
b. `30 g of C_(2) H_(6) = 54 g of H_(2) O`
`3 g of C_(2) H_(6) = 5.4 of H_(2) O`
c. `30 g of C_(6) H_(6) = 112 g of O_(2)`
`3 g of C_(2) H_(6) = 11.2 g of O_(2)`
weight of reactant `= 3 + 11.2 = 14.2 g`
Weight of product `= 8.8 + 5.4 = 14.2 g`
Hence, the law of conversation of mass is verified.
Second method:
Weight of `C` and `H` contained in `8.8 g`of `CO_(2)` and `5.4 g` of `H_(2) O` are
Weight of `C` in `8.8 g of CO_(2) = (12 xx 8.8)/(44) = 2.4 g`
Weight of `H` in `H_(2) O = (2 xx 5.4)/(18) = 0.45 g`
Total weight of `C` and `H` in the hydrocarbon after combustion `= 2.4 + 0.6 = 3 g`
Since the weight of `C` and `H` after combustion is the same as the weight of hydrocarbon before burining, the result are in accordance with the law of conservation of mass.