00:01
Let's take a look at the synthesis of methanol from carbon monoxide and hydrogen.
00:06
And let's say we start with 74 and a half grams of carbon monoxide and 12 grams of hydrogen.
00:19
Let's figure out what the limiting reactant is.
00:22
So to figure out limiting reactant, we should go to our product and we should choose the one that gives us fewer moles.
00:29
And so let's go ahead and start with our carbon monoxide.
00:38
We have to convert to grams using molar mass, and the molar mass of carbon monoxide is 28 .01 grams.
00:52
Then we want to go to methanol, and we create one mole of methanol for every one mole of carbon monoxide we use, getting them from the coefficients of our balanced equation.
01:08
And we multiply, i'm sorry, we'll stop right here.
01:13
And when we stop right here, we're going to find out how many moles we would make of methanol.
01:21
And it turns out that we would make 2 .66 moles.
01:29
We're going to do the same thing with our hydrogen.
01:32
We have 12 grams.
01:41
And the molar mass of h2 is 2 .016 grams.
01:46
Mole ratio here for every one mole with methanol we make, we have to use, look at the coefficient, two moles of hydrogen, and when you do the math on that, you find you get 2 .98 moles of methanol.
02:07
So the lower value is 2 .66.
02:10
That implies that co is the limiting reactant.
02:16
So carbon monoxide is our limiting reactant.
02:21
And let's say we wanted to figure out how much or how many grams of the excess reactant was left over...