00:01
Okay, so here you want to calculate the mass of water produced when 5 .72 grams of butane reacts with excess oxygen.
00:07
So first we need the balanced equation for butane.
00:11
So c4h10 plus o2 goes to co2 and water.
00:21
Now here, if we're going to go ahead and balance this, we can go ahead and balance the carbons on each side and the hydrogens.
00:29
So we have four carbons and 10 hydrogens on each side.
00:33
Then to balance the oxygens, we can balance the oxygens, me you can see that we have 13 on the right side.
00:38
And on the left side, the only thing with oxygen in it is this o2.
00:42
So in order to balance that, o2 can only balance even numbers.
00:47
We're right now on the right side we have an odd number.
00:49
So if you ever run into an even odd problem like this, what you can do is just double all your existing coefficients and then continue from there.
00:56
So to double all of the existing ones, this turns into a two, this turns into an eight, and this turns into a 10.
01:06
So then to balance the o2, we now have, on the right side, we have 26 oxygens, which means we'll need a 13 here.
01:14
So we have 5 .72 grams of butane, and we want it to get into grams of water...