Tue October 16, 2012 By:

0.44g of a hyrocarbon on compete combustion with oxygen gave 0.88g of water. show that the results are in aggrement with law of conservation of mass.

Expert Reply
Thu October 18, 2012
Given that, CnH2n+2 = 0.44 g of a hydrocarbon.
CnH2n+2  + O2 -------> nCO2 + (n+ 1)H2O

CO2 produced = 0.88 g 
We know, 44 g CO2 contains 12 g C
then 0.88 g CO2 will contain 0.88 x 12/44 = 0.24 g C

Also, 18 g water contains 2 g of H
then 1.8 g water will contain 1.8 x 2/18 = 0.2 g of H

Remember for burning, the O2 was in excess which can be ignored. Since the hydrocarbon contains only C and H, 0.44 g of C and H in the products.
 Thus, all of C and H from the reactant are conserved in the product.
Related Questions
Thu September 14, 2017

What is a mole?

Home Work Help