Balance each of the following chemical equations. a. Pb(NO3)2(a q)+K2 CrO4(a q) →PbCrO4(s)+KNO3(

Balance each of the following chemical equations. a....

Balance each of the following chemical equations. a. $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{2} \mathrm{CrO}_{4}(a q) \rightarrow$ $\mathrm{PbCrO}_{4}(s)+\mathrm{KNO}_{3}(a q)$ b. $\mathrm{BaCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{BaSO}_{4}(s)+\mathrm{NaCl}(a q)$ c. $\mathrm{CH}_{3} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ d. $\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{S}(s)+\mathrm{SO}_{2}(g) \rightarrow$ $\mathrm{CO}_{2}(g)+\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(a q)$ e. $\mathrm{Cu}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow$ $\operatorname{CuSO}_{4}(a q)+\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$ f. $\operatorname{MnO}_{2}(s)+\operatorname{HCl}(a q) \rightarrow \operatorname{MnCl}_{2}(a q)+\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$ g. $\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{KI}(a q)+\mathrm{HCl}(a q) \rightarrow$ $\mathrm{AsI}_{3}(s)+\mathrm{KCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ h. $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(a q)+\mathrm{I}_{2}(a q) \rightarrow \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}(a q)+\mathrm{NaI}(a q)$

Balance each of the following chemical equations.
a. $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{2} \mathrm{CrO}_{4}(a q) \rightarrow$
$\mathrm{PbCrO}_{4}(s)+\mathrm{KNO}_{3}(a q)$
b. $\mathrm{BaCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{BaSO}_{4}(s)+\mathrm{NaCl}(a q)$
c. $\mathrm{CH}_{3} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$
d. $\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{S}(s)+\mathrm{SO}_{2}(g) \rightarrow$
$\mathrm{CO}_{2}(g)+\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(a q)$
e. $\mathrm{Cu}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow$
$\operatorname{CuSO}_{4}(a q)+\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$
f. $\operatorname{MnO}_{2}(s)+\operatorname{HCl}(a q) \rightarrow \operatorname{MnCl}_{2}(a q)+\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$
g. $\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{KI}(a q)+\mathrm{HCl}(a q) \rightarrow$
$\mathrm{AsI}_{3}(s)+\mathrm{KCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$
h. $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(a q)+\mathrm{I}_{2}(a q) \rightarrow \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}(a q)+\mathrm{NaI}(a q)$

Key Concepts

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions involve the transfer of electrons between species. Balancing these equations may require additional steps, such as the half-reaction method, to account for both mass and charge conservation, making them a specialized application of balancing chemical equations.

Law of Conservation of Mass

This fundamental principle states that matter cannot be created or destroyed in a chemical reaction. It requires that the number of atoms for each element is identical on the reactant side and the product side, which is the basis for balancing chemical equations.

Stoichiometry

Stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. Balancing equations uses stoichiometric coefficients to ensure that the proportions of substances conform to the conservation of mass, allowing for the calculation of yield, excess reactants, and other important reaction parameters.

Balancing Chemical Equations

Balancing chemical equations is the process of inserting appropriate coefficients before compounds in a reaction to ensure that the number of atoms for each element is equal on both sides of the equation. This process is crucial for accurately representing and predicting the outcomes of chemical reactions.

Chemical Equation Coefficients

Coefficients in a chemical equation indicate the relative number of moles of each substance involved in the reaction. They are adjusted during the balancing process to achieve an equal number of atoms for every element on both the reactant and product sides, reflecting the stoichiometry of the reaction.

Transcript

00:01 Today in this video, we're going to talk, we're going to balance some equations.

00:07 So balance equations, you must have the same number of atoms on both sides.

00:14 So on the side of the left, we have the reactants, and on the right we have the products.

00:22 So we want to make sure that whatever you have on the reactant side is you have the same atom, number of atoms as you have on the product side.

00:30 A few things i'm going to go over before we get into actually balancing the equation or the substantiated.

00:34 Scripts.

00:35 Now, if you look at the first reactant, you have pb parentheses with no3 with a two next to it.

00:43 So the three under, that's next to the o is for oxygen.

00:49 So you have three oxygens bonded to one nitrogen.

00:54 And you can see the same thing with the next reactant.

00:56 You have two potassium bonded to a chromate.

01:01 And you see you have an oxygen, four oxygen.

01:05 That are bonded to one chromium.

01:08 Same thing with the other one.

01:09 You have potassium with a nitrogen, and then you have a 03, which means you have three oxygens.

01:17 Okay, so it's telling you how many atoms you have in the molecule.

01:23 And if you look at the first reactant again, we see the parentheses around the nitrate.

01:32 And what that means is do you have two nitrates bonded to the lead? so just basically when you have the subscripts, it's just telling you that there's more than one atom.

01:46 And how many of that? and there's more than one in the molecule.

01:52 And if you don't see a subscript, that means it's a one.

01:54 One is implied.

01:56 Same thing goes for coefficients.

01:57 As the equation is written right now, you see there's no coefficients.

02:05 So it's assumed that you have like a one here.

02:10 That's unwritten.

02:11 So as written, it's assumed you have one mole of the first reactant bonding with one mole of the second, yielding one mold of the first product plus one mold of the last one.

02:28 But if you were to look at how many atoms on both sides, we'll find that it's not necessarily going to be the same until we inspect it to make sure it's properly balanced.

02:42 So we'll start with the lead.

02:44 If you look on the left side, you'll see you have one lead atom, and then in the product side you have one lead.

02:51 So that's balance, and that's okay.

02:54 Next, let's look at nitrate.

02:57 You see this side you have two nitrates, but when we look on the reactant side, i only see one nitrate.

03:06 So to change that, what we would do is put a two in front of it.

03:14 Now we have two moles of potassium nitrate.

03:17 Instead of just one.

03:24 So now we have two of them.

03:29 That's balanced.

03:30 But when we have two, now we have two potassium.

03:32 Then we look over at the product side, i'm sorry, the reacting side.

03:36 We see there's two potassiums.

03:39 So we have two potassium on both sides.

03:41 That's balanced.

03:42 And then look at the chromates, and you see that there's one chromate on both sides.

03:50 So this is balanced.

03:52 So the only thing we needed to do to balance this equation was they added to on the reactive side.

04:00 So let's go on to part b.

04:02 We have ba.

04:24 Okay, so for part b, we do the same thing.

04:27 You just inspect it.

04:28 So i'll start with barium.

04:29 So you have one barium on the reaction side.

04:32 Go over the products.

04:33 We see one.

04:35 And you see ion, like, sulfate.

04:39 You can say, okay, let's look at it as a unit.

04:41 So there's one sulfate on the side, on the product side, one sulfate on the right.

04:46 React inside.

04:49 And then we'll look over on the product side again, you see that there's two chlorides.

04:59 But when we look at the chloride in the salt, table salt molecule, you see that there's only one.

05:09 And then you look on the product, you see again, there's two sodiums and only one sodium.

05:15 So to fix it out, all we need to do is put in a two.

05:18 Now it's balanced.

05:19 Now there's two sodium, two chlorides, two chlorides, two sodiums, and then one of everything else.

05:26 So that one is balanced.

05:30 Okay, so for the next one for part c, we'll do inspection again.

05:35 So we'll start off with the carbon.

05:37 You'll see there's one carbon, the alcohol, then you look over the co2, one carbon, balance.

05:46 I just look at the oxygen, i'm sorry, not the oxygen, the hydrogen first.

05:50 So we have three hydrogen and one more makes four on this side but then we look at water only has two hydrogens so to balance the hydrogens we could put a two in front of it and that would give us two uh what um two h2 o's and then for the oxygens so let me put the two in right now so on this side we're going to have two oxygens plus two more which be two oxygens on the product side.

06:26 On the reacting side, though, that leaves us with three.

06:29 We have two in the oxygen molecule plus one in the alcohol.

06:36 So you have three oxygens on the side and four on this side.

06:42 Now to balance that, if we want to make it so that it's the same, we can put a one and a half in front of the oxygen and then you'd have four on both sides.

06:58 And then everything we balance.

07:02 The only thing is, usually we don't see fractions or decimals when we have equations.

07:10 People like to use whole numbers.

07:16 So what you can do to do that is multiply the whole thing by 2.

07:23 So you end up with, so we do multiply by 2.

07:30 And then distribute it, you end up with 2.

07:41 Plus two times one and a half it would be three and two co2 plus four okay so for part d do the same thing we'll start off with the first reactant you have two sodium and so we look over on the product side and we have two sodium as well and then we see there's one carbon and then in the co2 has one carbon let's add up all the oxygen.

08:31 So we have three oxygen here and then two more, which would be five oxygen.

08:37 And then on the reactants, we have the co2 has two.

08:42 And then we have three...

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