10 ml of gaseous hydrocarbon was fully burned in 50 ml of oxygen, and liquid water and 35 ml of mixed gas were obtained I figured out that there are 6 h, so how can I determine the number of C The answer is C2H6 and C3H6 I have C4H6, too. C2H6 and C3H6

Let CxHy, complete combustion
CxHy + (x + Y / 4) O2 = xco2 + Y / 2H2O (L), the gas volume decreases
1--------(x+y/4)------x-------------------(1+y/4)
10ml ------ 50ml excess -------- (60-35)
The solution is y = 6
That is to say, the hydrocarbon meets cxh6
Because the number of C atoms can not be 1, the number of H atoms is more than saturated, and because it is a gaseous hydrocarbon, the number of C atoms is less than or equal to 4
Therefore, it can be C2H6, C3H6, C4H6

10ml of a gaseous hydrocarbon is burned in 50ml of oxygen to obtain liquid water and 35ml of gas
1.CH4
2.C4H8
3.C3H8
4.C3H6
I chose D, right
I'll think about it again

I will calculate by volume ratio, which is equal to molar ratio A. and oxygen full reaction ratio 1:2b. And oxygen full reaction ratio 1:6 C. and oxygen full reaction ratio 1:5 D. and oxygen full reaction ratio 1:4.5. Because it is gaseous hydrocarbon, and the ratio of oxygen is 1:5, it can be considered

A 10 ml gaseous hydrocarbon is fully burned in 50 ml oxygen to obtain 35 ml gas and water
Understand

Hydrocarbon: oxygen: CO2 = 2:10:7, that molecule of hydrocarbon contains 3.5 C, oxygen consumption is C number + H number / 4, one molecule of hydrocarbon consumes 5 molecules of oxygen, 3.5 out to C, the remaining 1.5 is h, so h number is 1.5 * 4 = 6, this thing is c3.5h6, that is c7h12, it should be methylbenzene, but toluene is not a gas under normal temperature and pressure

10 ml of gaseous hydrocarbon was fully burned in 50 ml of oxygen, and liquid water and 35 ml of mixed gas were obtained I figured out that there are 6 h, so how can I determine the number of C The answer is C2H6 and C3H6 I have C4H6, too. C2H6 and C3H6