2CnH2n+2+3n+1 O2= 2n CO2+2n+2H2O
1.we find the amount of oxygen that reacts
75-12.5=62.5sm3
2.we can determine the proportion of hydrocarbons
2mol hydrocarbons 3n+1 oxygen
12.5sm3 hydrocarbons 62.5sm3 oxygen
we get the value of n from the proportion
3n+1=62.5*2/12.5
3n+1=10
3n=9
n=3
hence the formula of the hydrocarbon is C3H8
3.we calculate the molar mass of C3H8
12*3+1*8=44g/mol
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