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0.44 g of a hydrocarbon on complete comb...

0.44 g of a hydrocarbon on complete combustion with oxygen gave 1.8g water and 0.88 g carbon dioxide. Show that these resutls are in asccordance with the law of conservation of mass.

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A hydrocarbon is a compound which consists of carbon and hydrogen only. It undergoes combustion forming carbon dioxide and water as products.
Formula of Carbon dioxide=`CO_(2)`,
Molecular mass `=12+32=44g`.
Formula of water `=H_(2)O`.
Molecular mass =2+16=18g
Mass of carbon in 0.88 g of `CO_(2)=(12)/(44)xx0.88=0.24g`
Mass of hydrogen in 1.8 g of `H_(2)O=(12)/(44)xx1.8=0.20g`
Mass of hydrogen in 1.8g of `H_(2)O=(2)/(18)=0.20g`
Total masses of carbon and hydrogen in the products
`=0.24+0.20=0.44g`
This is equal to the mass of hydrocarbon before combustion.
Thus, the results are in accordance with the law of conservation of mass.
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Knowledge Check

  • 0.58 g of hydrocarbon on combustion gave 0.9 g water. The percentage of carbon is about

    A
    75.8
    B
    82.7
    C
    27.85
    D
    68.8

  • 3 g of a hydrocarbon on combustion with excess of oxygen produces 8.8 g of CO_2 and 5.4 g of H_2 O . The data illustrates the law of

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