Write the balanced equation then outline the steps necessary...

Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following: (a) The number of moles and the mass of chlorine, $\mathrm{Cl}_{2},$ required to react with 10.0 $\mathrm{g}$ of sodium metal, Na, to produce sodium chloride, NaCl. (b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 $\mathrm{g}$ of mercury(II) oxide. (c) The number of moles and the mass of sodium nitrate, required to produce 128 $\mathrm{g}$ of oxygen. (\textrm{NaNO} _ { 2 } \text { is } the other product.) (d) The number of moles and the mass of carbon dioxide formed by the combustion of 20.0 $\mathrm{kg}$ of carbon in an excess of oxygen. (e) The number of moles and the mass of copper(II) carbonate needed to produce 1.500 $\mathrm{kg}$ of copper(II) oxide. $\left(\mathrm{CO}_{2}\right.$ is the other product.) (f)

Transcript

00:01 So in this problem, we need to explain how to convert from grams of one substance into moles and then grams of another using a balanced chemical equation.

00:12 The first one they give us, we have sodium plus chlorine.

00:19 Remember, chlorine is diatomic, so it's cl2, producing an acl in a synthesis reaction.

00:26 They tell us the mass of the sodium, and we need to convert.

00:33 Into moles and then grams of cl2.

00:37 Before you do anything, you have to balance this equation, the only thing we need to do is to put a two in front of the nacl and then a two in front of the n .a.

00:47 That will let us balance that equation.

00:50 So we're in grams of n .a.

00:54 In order to get to moles of n .a., we're going to have to divide by the molar mass, and i'm going to use this a lot, so i'm going to refer to this as capital mm.

01:10 When i write it from now on, we're going to divide by the molar mass of n .a.

01:18 To get to moles.

01:22 From there, we need to use coefficients.

01:31 Those are from our balanced chemical equation in order to convert from moles of n .a.

01:41 Two moles of the cl2.

01:50 That answers the first part of the question.

01:52 From here, i'm going to multiply by the molar mass of cl2 in order to get to grams.

02:02 You will see as we go, these are all going to follow this same basic pattern.

02:08 In b, we have mercury 2 oxide, and that is going to decompose.

02:17 It's going to break down into hg plus o2.

02:21 Remember, oxygen is diatomic.

02:24 Balance it out by putting twos in front of both mercury -containing compounds.

02:33 So first thing we're going to do, they gave us the mass of hgo.

02:39 So i need to divide the mass of hgo by its molar mass.

02:53 That gets me to moles.

02:55 Okay.

02:56 After i'm in moles, i'm going to use coefficients.

03:06 This will allow me to convert from moles of hgo to moles of o2.

03:19 And lastly, i'm going to multiply by the molar mass of o2 to get that into grams.

03:27 Same exact pattern as the one above it.

03:30 This is going to be the way you have to manipulate all of these problems in order to solve them.

03:38 The next one, we have n -a -n -o -3, decomposing to form n -a -n -o -2, and oxygen.

03:50 We have four oxygens on the product side.

03:52 We have three on the reactant side.

03:54 You need to get this to be even, because otherwise the three is never going to work with the even on the other side.

04:04 So start by putting a two there.

04:06 That gives me six oxygens...

Question

Please give Ace some feedback