00:01
In this equation, write the balanced equation for stoichiometric combustion of dot can c12h26.
00:08
The combustion reaction of dot can with oxygen can be written as c12h26 plus o2 co2 plus h2.
00:17
Now we have to balance the equation that is c12h26 plus 37 by 2 o2 c12co2 plus 13 h2o.
00:26
Now step 2, we have to calculate the mass air to fuel ratio for the stoichiometric combustion of dot can.
00:35
The mass of dot can is given by the molecular weight 12 into 12 gram per mole, carbon plus 26 into 1 gram per mole, hydrogen 170 gram per mole.
00:45
The mass of oxygen required for stoichiometric combustion is 37 by 2 into 16 gram per mole.
00:51
Oxygen is 296 gram.
00:53
Since air is 21 % oxygen and 79 % nitrogen, we can calculate the mass of air required 296 gram per mole divided by 0 .21 is equal to 1409 .52 gram per mole.
01:11
Now we can find the mass air to fuel ratio.
01:15
Mass air to fuel ratio is 1409 .52 gram per mole divided by 170 gram per mole is equal to 8 .29.
01:48
Now we have to calculate the air fuel equivalence ratio.
02:08
So given mass of fuel is 2 .5 gram...