posted by Amy~ .
A mixture of methane (CH4) and ethane (C2H6) of mass 13.43g is completely burned in oxygen.
If the total mass of CO2 and H2O produced is 64.84 g, calculate the fraction of CH4 in the mixture.
CH4 + C2H6 arrow CO2 + H2O
Would I do 13.43g/16.04g CH4 ?
Start with the chemical equations of the complete combustion of CH4 and C2H6,
CH4 + mO2 -> CO2 + 2H2O ....(1)
C2H6 + nO2 -> 2CO2 + 3H2O ....(2)
Calculate the total mass of CO2+H2O produced if all 13.43g of gas were CH4.
Do the same for C2H6.
Find the ratio of each gas by proportion.
I don't get this.
Would I solve for m
13.43 + m32 = 36.04
Also for CO2 + 2H2O ....(1)
2CO2 + 3H2O ....(2)
why does the bottom one have 2CO2 and 3H2O?
First, you don't need to know m and n, because only the products count. However, m=2 and n=3.5.
If written properly, equation 2 should be all multiplied by 2.
For equation 1, calculate the RMM for each ingredient, then calculate the ratio of masses of CH4 and the products, namely CO2 and 2H2O.
Thus (12+4)=16 g of CH4 will produce (12+32 + 2(2+16))=80 g of CO2 and H2O combined. The ratio of mass of products/mass of CH4 is therefore 80/16=5.
Calculate similarly the ratio of mass of products / mass of C2H6, call it x.
The ratio of mass of products / mass of mixture = 64.84/13.43 = 4.828
From the ratios 5 and x, calculate the mix of CH4 and C2H6 to give a ratio of 4.828.
The explanation is poor try putting up the answer ! So people can see the way to proably do it!