00:01
Good day in this question we are asked to solve for the percent yield of co2 in this reaction.
00:05
So first we need to write a balanced equation between pentane and oxygen, which is c5h12, plus o2 that will yield co2 and h2.
00:19
Balancing the equation, we have 5, co2 and 6h2 plus 10.
00:31
So we have 8 ,000, 6, 06 plus 10.
00:32
From this we need to solve for mole of bentane and oxygen first from the given mass so mole of bentane is equals to from the given mass five grams divided by molar mass of 72 .15 grams per mole this is equals to 0 .0693 mole next mole of o two is equals to the given mass which is 5 grams also divided by 32 grams per mole.
01:13
This is equals to 0 .156 mole.
01:18
Next way then defy which is the limiting reactant.
01:21
By solving mole of pentane needed to react from 02, substituting will 0 .156 mole of 02 times 1 mole of pentane is 8 moles of o2 this is equals to 0 .0 .0 .0 .0 .0.
01:44
From the 0 .195 mole of pent...