00:01
In this question, considering 1 kilomol fuel the composition equation can be written as 0 .65 ch4 plus 0 .08 h2 plus 0 .18, h2 plus 0 .18 and 2 plus 0 .13, o2 plus 0 .06, co2 plus a theoretical, multiply o2 plus 3 .76 n2.
00:46
This will give us x, carbon dioxide, plus y, water, plus z, nitrogen.
01:00
So the unknown coefficient in the above equation can be determined from the mass balance.
01:07
So by balancing the carbon we will get 0 .65 plus 0 .06 equal x.
01:18
So in this case x equal 0 .71.
01:22
By balancing hydrogen we have 0 .65 multiply 4 plus 0 .08 multiply 2 equals to y so y in this case equal 1 .38 by balancing oxygen so we have 0 .03 plus 0 .06 plus a theoretical equals x plus y over 2 so from this equation a theoretical equals a theoretical equals x plus y over 2 so from this equation a theoretical equals 1 .31 and and finally by balancing nitrogen, we have 0 .18 plus 3 .76.
02:19
A theoretical equals z or z, so z equals 5 .106...