00:01
Here in this problem, we have to consider this equation.
00:05
If 2 .5 grams n .a react with 1 .98 grams of o2, then we have to calculate the theoretical yield of sodium oxide.
00:15
First, we will calculate the moles of n .a.
00:19
And moles of o2.
00:22
Moles of n .a.
00:23
That will be 2 .55 grams n .a.
00:27
Given, divided by the molar mass of n .a.
00:30
22 .9 .9.
00:31
Grams per mole and we get 0 .111 most now we'll find out moles of o2 and that will be mass of o2 given here divided by the molar mass of o2 32 grams per mole and we get 0 .0 .062 now we'll find out which one is limiting reactant.
01:01
So let's let's find out here 0 .111 mole n .a requires how many moles of o2? that will be 0 .111 mole n .a times conversion factor 1 mole o2 divided by 4 moles na.
01:26
Because according to this equation given 4 moles na require 1 mole of o2 and 2 of n .a .2 .o.
01:37
Are produced.
01:38
So here we get 0 .02775 moles o2...