00:01
In this question we are assigned to the task to calculate the percentage yield of the reaction.
00:08
It is given that we have 124 .0 gram of lithium, 98 .2 gram of nitrogen gas, that is n2, and 195 gram of l .i .3n, that is, lithium nitrite.
00:27
The balanced chemical equation is 6 l .i solid plus n2 gas gives out 2 l .3n solid.
00:40
Now, the molar mass for lithium is 6 .941 gram per more and the molar mass for n2 is 28 .02 gram per more.
00:55
Therefore, moles of lithium is equals to mass divided by molar mass, that is 124 .0 gram divided by 6 .941 gram per mole, which comes out to be 17 .8 6 .8 6 .6 mall.
01:14
Similarly, the malls of n2 is equal to 98 .2 gram divided by 28 .02 gram divided by 28 .02 per mole, and it comes.
01:25
Comes out to be 3 .50 mall.
01:29
Now, according to the balanced chemical equation, 6 malls of l .i will react with 1 mole of n2...