00:01
Okay, so for this question, they want us to calculate the percent yield of nhco3.
00:06
So in their experiment, they made 66 kilograms or 66 ,000 grams.
00:13
And this is the chemical equation that they use.
00:18
So the first thing we have to do is we started off with 58 .5 kilograms of nacl and 18 .8 kilograms of nh3.
00:26
So we have to find which one is the limiting reagent.
00:29
So you can pick nacl or nh3.
00:32
I'm going to pick nacl.
00:34
I converted the 58 .5 kilograms to grams, so that is 58 ,500 grams of nacl.
00:44
So one mo of nacl is 58 .5 grams.
00:52
Now it's time to do the multiple ratio of nacl or nh3.
00:56
So we have one mole of nacl, one mole here, is needed to to react with one mole of h3.
01:08
So one mole of nh3 is equal to 17 grams.
01:18
Right, so using this, we see that we get 17 ,000 grams of nh3.
01:25
So this is the amount of nh3 that we need to react with 58 ,500 grams of nacl.
01:33
And we see that we started off with 18 ,800 grams of nh3.
01:38
So if we only need 17 ,000 of it, that means we'll still have an h3 left, meaning that nh3 is not the limiting reagent and nacl is a living reagent...