00:10
We are asked relating to this reaction, how many moles of ammonia can be produced from 24 .0 moles of hydrogen and excess, which i will now breathe, n2, and we're told to go to three significant figures, which i would have anyway.
00:42
24 .0 moles of h2, converting this to moles of ammonia, we will use our balanced chemical equation to obtain the mole ratios.
01:00
And this will be, i believe, 16.
01:06
16 .0 moles of nh3.
01:12
2.
01:18
That's 1.
01:19
2.
01:20
We're asked to calculate the mass of water when 1 .05 grams of butane reacts with excess oxygen.
01:44
Okay, so butane is c4h10.
01:50
Can't believe they didn't give that to you.
01:54
And that reacts with oxygen to produce co2 plus h2o.
02:03
I'm going to put a 10 there and a 2 here and an 8 here.
02:08
And i've got 16 plus 10 is 26 and a 13.
02:13
Okay, so there's my balanced chemical equation.
02:17
We're also going to go to three sigphakes with the correct units.
02:20
1 .05 grams times 1 .05 grams.
02:36
Let me get to my molar mass of c4h10.
02:43
Butane will be 58 .12.
02:58
Then i'm going to use my mole ratio for water in my numerator and butane in my denominator.
03:12
And we'll multiply that by the molar mass of water, which is 18 .02 grams per mole.
03:22
Let me put these numbers into my calculator.
03:33
And this will equal 1 .63 grams of h2o.
03:41
1 .05 times 10 times 18 divided by 58 divided by 2.
03:51
Okay.
03:54
Now i've got more to do here.
03:57
Number three.
04:03
Calculate the mass of butane needed for 89 grams of co2.
04:17
89 .0 grams of co2.
04:21
Okay, this one's pretty easy as well...