00:02
Here in this problem, given this reaction, 4 aluminum plus 3 .02 produces 2 aluminum oxide.
00:11
Here, given 40 gram of aluminum and 19 gram of oxygen for the reaction.
00:17
We have to find out more of aluminum oxide form.
00:21
Now, let's find out 40 gram aluminum means how many more? 40 gram aluminum that is 40 gram divided by the molar mass of aluminum that is 26 .98 gram.
00:44
So we get 1 .4825 more.
00:50
Now we have 19 gram of oxygen.
00:54
Let's find out how many modes that one is.
00:57
So 19 gram oxygen, that is 19 gram divided by the molar mass of oxygen gas, that is 32 gram.
01:12
And we get 0 .59375 more.
01:21
Here, from the reaction, we see 3 mole oxygen needs 4 moles of aluminum for the reaction.
01:30
So let's write 3 more oxygen requires 4 more of aluminum.
01:44
Therefore we have 0 .59375 more oxygen.
01:51
So 0 .59375 more o2 requires 4 more 4 more .4 more.
02:04
Aluminum divided by three more oxygen times 0 .59375 more o2...