00:01
So in this question the equation is given to us when the weight of aluminum is also given that is 91 .41 gram and the percentage yield that is 46 .32 % what we have to find out? we have to find out the actual weight of aluminium with the percentage yield that is 46 .32%.
00:16
So starting on the given condition that one thing is also given to us that co, cl2, water is also present in excess we can see the present in large amount so from the balance equation is clear that if there are two moles of aluminium two moles of aluminium then there will be two moles of alcl 3 will be produced two moles of alc l3 definitely would be produced which is clear from the equation view so it is clear if two more of aluminium there is then 2 mole of alcl3 so quite clear that if there is one mole of aluminium and definitely one mole of alcl3 would be produced.
01:13
Now further what you have to do here we can find out the moles using for molar that number of moles should be equal to number of moles n that's for aluminium should be equal to given mass given mass divided by molar mass given mass of aluminum is given that is equal to 90 1 .41 molar mass aluminum is that should be equal to 27 gram per more so number of moles should be equal to comes out to number of moles of aluminium that is equal to 3 .385 3 .385 moles of aluminium is here.
01:59
Now we know that the 1 mole of aluminium it will produce 1 mol of alcl3 definitely if the number of moles of aluminium is 3 .385 it is quite much clear that number of moles of alcl3 should also be equal to 3 .385.
02:17
Now again by using the formula that number of moles should be equal to given mass upon molar mass is clear that mass of alcl3 should be equal to number of moles of alcl3 multiply the molar mass of alcl3, molar mass of alcl3.
02:36
So number of moles should be equal to that is 3 .385, 3 .385 and the motor amounts of alcl3 is 133 gram per mole.
02:49
From here, the mass of alcl3 would be equal to 450 .2 gram...