f.6 的CHEM 問題

2007-09-24 3:23 am

48 CM3 of a mixture of methane and ehtane were exploded with 180cm3of oxygen. Fter cooling to the original room temperature, the volume of the residaul gas was noted. It was found to decreaseby64 cm3 when treated with potassium hydroxide solution. Assuming constant pressure throughout, calculate the composition of the mixture by volume.

回答 (1)

Carson

2007-09-24 4:32 am

✔ 最佳答案

48 cm3 of a mixture of methane and ethane
180 cm3 of oxygen
CH4 + 2 O2 ---> CO2 + 2 H2O
vol of CH4 : O2 = 1 : 2
vol of CH4 : CO2 = 1 : 1
2 C2H6 + 7 O2 ---> 4 CO2 + 6 H2O
vol of C2H6 : O2 = 2 : 7 or 1 : 3.5
vol of CH4 : CO2 = 1 : 2
KOH + CO2 ---> KHCO3
as residaul gas decrease by 64 cm3, so vol of CO2 is 64 cm3
First we assume O2 is in excess, and there is x cm3 of CH4 and 48 - x cm3 of C2H6
so vol of CO2 formed is x + 2(48 - x) = 64
96 - x = 64
x = 32
there is 32 cm3 of CH4 and 16 cm3 of C2H6
vol of O2 consumed is 2 (32) + 3.5 (16) cm3
= 120 cm3
so the oxygen is in excess is true
the composition of the mixture by volume is 32 cm3 of CH4 and 16 cm3 of C2H6

I hope this can help your understanding. =)

參考: My knowledge

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