00:01
So we have this question that the combustion reaction of glucose given to find out the maximum of 0 here consume.
00:05
And we also find out the limiting reagent and the remaining glucose left in the reaction.
00:11
First of all, starting with the given values that we know that mass of glucose should be equal to 180 gram per more.
00:19
And mass of oxygen that would be empty equal to 32 gram per more.
00:23
And marks of co2 let it be m3 should be equal to 44 gram per more.
00:28
So now i find out here that the number of moles of oxygen is 6 as 0 06 in the balance equation.
00:35
So weight of 6 mole of oxygen should be equal to weight of 6 moles of oxygen should be equal to 6 into 32 that which should be equal to 1 .92 grams.
00:55
Similarly, the weight of 6 moles of co2 should be equal to weight of 6 moles of co2 should be equal to it should be equal to 140 or it should be equal to 64 gram.
01:17
It should be equal to 264 gram here.
01:22
Now, on the balancing equation is clear that 192 gram of season produce, 190, 2 gram of season produce.
01:36
How much value of co2? that 264 gram of co2.
01:44
So as for the demand of question here, what you have to do? 11 .7 gram will be quiet, will produce 11 .7 gram of oxygen will produce how much co2 of oxygen will produce that it should be equal to to 64 divided by 192 multiply with 11 .7 and max amount of co2 should be equal to, that it should be equal to 16 .0 by 75 grams...