7.82 g of methane, CH4, reacts with an excess of oxygen in...

7.82 g of methane, CH4, reacts with an excess of oxygen in the following unbalanced reaction. Balance the reaction and calculate the mass of water produced. Answer all questions using the dropdown menus for a-e CH4(g) + O2(g) ? CO2(g) + H2O(g) 1 CH4 (g) + 2 O2 (g) ? 1 CO2 (g) + [Select] H2O (g) [ Select ] 17.6g 0.974g 7.82g impossible to know. How many grams of water are produced?

Transcript

00:02 Hi there.

00:03 In this question, we are given this unbalanced equation.

00:06 I have emitted the states of matter because they are not important for what we are using this equation for.

00:11 And what we're using this equation for, the first thing is to balance it.

00:16 So balancing means we need the same number of each type of atom on both sides of the arrow.

00:21 So starting off, we see that there's one carbon on each side.

00:24 So that is balanced.

00:25 There are four hydrogen on the reactant side and only two on the product side.

00:29 That means in balancing, all we add are co -eastern.

00:32 We can't change subscripts.

00:35 So i need to put a coefficient of two to give me two times two or four hydrogen.

00:40 This gives me a total of two plus two times one or four oxygen on the product side.

00:48 And on the reactant side, we only have two.

00:50 So i need a coefficient of two in front of the oxygen.

00:54 All right.

00:55 This equation is now balanced.

00:59 And so the coefficients are one, two, one.

01:04 N2.

01:07 All right.

01:07 On to our next step then.

01:09 What we want to know is how much water is going to be produced when we start with 7 .82 grams of methane.

01:19 I apologize for my erasing here.

01:22 The eraser is very thin now in the new system and i don't know a way to make it fatter to erase efficiently.

01:31 All right.

01:32 So let's go on to that second part of the question then.

01:36 We have 7 .82 grams of the methane and excess oxygen.

01:43 And we want to calculate how many grams of water are produced.

01:48 So what we see here is that this is a mass to mass stoichiometry problem because we're starting with the mass of one thing in the equation and we're trying to determine the mass of something else in the equation.

02:02 So in any stoichiometry problem, in one of the steps we need to use the mole ratio.

02:08 That's how we convert from one substance to another.

02:11 And the mole ratio comes from the coefficients in this balanced equation.

02:15 That's why we needed a balanced equation.

02:18 We know in front of the ch4, since nothing's written there, it's a 1.

02:21 So we have a 1 to 2 mole ratio.

02:24 All right.

02:24 So what we're going to need to do is to convert from grams of methane to moles of methane...

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