00:01
Hi, so to solve for this, let us write first the balance equation for the reaction of methane and carbon tetrachloride.
00:10
So that's ch4, this is methane, and carbon tetrachloride is ccl4.
00:16
Then the product, we have ch2cl2.
00:21
So to balance this, we just need to write here 2.
00:25
Okay, so this is the balance equation.
00:27
Now we need to identify the limiting reactant and the excess reactant in this reaction.
00:34
So we'll solve first for the number of moles of each reactant.
00:38
We'll start with methane, number of moles of ch4.
00:42
If we have 7 .31, this is grams of methane.
00:47
Convert this to moles by dividing the molar mass of ch4.
00:50
That's 16 .05 grams here, and then moles in the numerator.
00:56
So that means we could cancel grams of ch4 and we'll get 0 .455.
01:04
This is moles of ch4.
01:07
And then for carbon tetrachloride, we have 29 .2 grams of ccl4.
01:17
Convert to moles by dividing the molar mass of ccl4.
01:21
That's going to be 153 .81 grams here, and then moles in the numerator.
01:27
So that means we could cancel grams and we'll get 0 .190.
01:34
This is moles of ccl4.
01:37
Now we will solve for the mole ratio.
01:42
So mole ratio in this case is the number of moles of the reactant divided by the coefficient of that reactant from the balance equation.
01:50
So for ch4, we have calculated 0 .455 moles divided by the coefficient of ch4 from the balance equation, which is 1.
02:02
So divided by 1, we'll still get 0 .455.
02:07
And then for ccl4, we have 0 .190.
02:13
This is moles divided by the coefficient of carbon tetrachloride, which is also 1...