Balance the following equations using the method outlined in...

Balance the following equations using the method outlined in Section 3.7 . (a) $\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2}$ (b) $\mathrm{KNO}_{3} \longrightarrow \mathrm{KNO}_{2}+\mathrm{O}_{2}$ (c) $\mathrm{NH}_{4} \mathrm{NO}_{3} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}$ (d) $\mathrm{NH}_{4} \mathrm{NO}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}$ (e) $\mathrm{NaHCO}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$ (f) $\mathrm{P}_{4} \mathrm{O}_{10}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}$ (g) $\mathrm{HCl}+\mathrm{CaCO}_{3} \longrightarrow \mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$ (h) $\mathrm{Al}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{H}_{2}$ (i) $\mathrm{CO}_{2}+\mathrm{KOH} \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}$ (j) $\mathrm{CH}_{4}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$ (k) $\mathrm{Be}_{2} \mathrm{C}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Be}(\mathrm{OH})_{2}+\mathrm{CH}_{4}$ (l) $\mathrm{Cu}+\mathrm{HNO}_{3} \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O}$ $(\mathrm{m}) \mathrm{S}+\mathrm{HNO}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O}$ (n) $\mathrm{NH}_{3}+\mathrm{CuO} \longrightarrow \mathrm{Cu}+\mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}$

Balance the following equations using the method outlined in Section 3.7 .
(a) $\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2}$
(b) $\mathrm{KNO}_{3} \longrightarrow \mathrm{KNO}_{2}+\mathrm{O}_{2}$
(c) $\mathrm{NH}_{4} \mathrm{NO}_{3} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}$
(d) $\mathrm{NH}_{4} \mathrm{NO}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}$
(e) $\mathrm{NaHCO}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$
(f) $\mathrm{P}_{4} \mathrm{O}_{10}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}$
(g) $\mathrm{HCl}+\mathrm{CaCO}_{3} \longrightarrow \mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$
(h) $\mathrm{Al}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{H}_{2}$
(i) $\mathrm{CO}_{2}+\mathrm{KOH} \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}$
(j) $\mathrm{CH}_{4}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$
(k) $\mathrm{Be}_{2} \mathrm{C}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Be}(\mathrm{OH})_{2}+\mathrm{CH}_{4}$
(l) $\mathrm{Cu}+\mathrm{HNO}_{3} \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O}$
$(\mathrm{m}) \mathrm{S}+\mathrm{HNO}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O}$
(n) $\mathrm{NH}_{3}+\mathrm{CuO} \longrightarrow \mathrm{Cu}+\mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}$

Transcript

00:02 Hi there.

00:03 In this question, we have a few chemical equations to balance.

00:07 And what we want to remember is balancing equations is based upon the law of conservation of matter.

00:14 Law of conservation of matter says that in any chemical reaction, you cannot create nor destroy matter.

00:22 So that means when we write an equation to represent that reaction, we have to make sure that however many atoms of each element we start with, we must also see that.

00:32 That same number and type of atom on the product side.

00:36 So therefore, we have to balance equations to make that happen.

00:40 So let's start off with letter a.

00:42 We have n205, decomposing to form n204, and oxygen gas.

00:58 Right.

00:58 So right now, we have the same type of atom on both sides of the arrow.

01:02 We have nitrogen and oxygen on both sides, but we have to check and make sure that the number of each is appropriate.

01:09 So on the left side, we have two nitrogen.

01:12 On the right side, we also have two nitrogen.

01:15 So nitrogen is good.

01:17 On the left side, we have five oxygen, but on the right side, we have four in the no4 and another two in the o2.

01:26 One of the techniques they showed you in the chapter was that i would like to have five oxygens on the right side so it would match the five on the left.

01:37 And i could do that by putting in a coefficient of one half here because one half times two gives me one and then i have five oxygens but we know coefficients cannot be it cannot be fractions.

01:52 They have to be whole numbers.

01:54 So think about math class and if you have a coefficient of one half and you're trying to get rid of that coefficient, you know that you can multiply by the reciprocal or in other words two.

02:06 So if i multiplied two times one half that would give me one but i would also have to multiply each of the other coefficients in the equation by two.

02:14 So let's go ahead and try that.

02:17 If i multiply one half by two, it becomes one, but i have to multiply my other coefficients, which are currently 1 times 2, which gives me 2.

02:27 So math tells me this should be balanced now, but let's check out chemistry -wise.

02:32 So the coefficient of 2 gets multiplied times the subscript, so 2 times 2 is 4, nitrogen.

02:38 On the product side, we have 2 times 2 or 4.

02:41 Nitrogen.

02:42 Going back to the reactant side, for oxygen, we have two times five or 10.

02:48 As we go to the product side, we see that we have two times four oxygen in the n204, and then we have two more oxygen in the o2.

03:01 So sure enough, that adds up to 10 oxygen atoms.

03:03 So this is now balanced.

03:05 This equation is now balanced as 2n205 yields 2n204 plus o2.

03:14 Let's go on to letter b.

03:17 Letter b is going to be very similar.

03:19 So i'm going to go through it a little more quickly.

03:22 We have kno3 and it reacts to form kno2 plus o2.

03:34 For the k or the potassium, there's one on each side.

03:38 There's one nitrogen on each side.

03:40 But the reactant side only has three oxygen whereby the product side has four.

03:45 Again, i noticed that if i could multiply that 02 by 1 half, then i would only have three oxygen on the product side.

03:54 But i know i can't have coefficients that are not whole numbers.

03:58 But if i multiply one half by two, then it just equals one.

04:03 But then i have to multiply my other coefficients through by 2 as well.

04:07 So this should now be balanced.

04:09 2k on each side, 2n on each side, 2 times 3, which is 6 oxygen on the reactant side, and two times two, which is four, plus two more oxygen on the product side.

04:24 So this one is now balanced.

04:28 All right, let's go on to letter c.

04:32 Letter c, we have ammonium nitrate forms n2o and h2o.

04:52 Okay, let's start with any element except oxygen.

04:55 We don't want to start with oxygen because it occurs in both of the products.

04:59 So we always want to save elements that occur in more than one reactant or product for last.

05:05 So starting with nitrogen.

05:08 Let's see.

05:09 We have one, two nitrogen on the reactant side.

05:14 There's one there, and there's another one there.

05:16 So make sure that you find all of them.

05:19 So there's two nitrogen.

05:21 And on the product side, we have two nitrogen.

05:25 On the reactant side, there are four hydrogen.

05:28 Product side, we only have h2.

05:30 So if we need four hydrogen, think what coefficient we would have to add, so then we multiply by the subscript to two, it gives us four.

05:39 Well, that coefficient is going to have to be a two, because two times two is four.

05:46 So now i have four hydrogen.

05:48 Let's check the oxygen.

05:50 Three oxygen on the reactant side, and we have one oxygen in the n2o, and two more oxygen because we have two molecules of water.

06:00 So 1 plus 2 gives us 3 oxygen.

06:04 So this equation is now balanced.

06:07 Going on to letter d.

06:10 Letter d, we have nh4, n -o2.

06:19 That is breaking down to form nitrogen, gas, and water.

06:29 All right, starting with the nitrogen again, there are two nitrogen.

06:33 There's one nitrogen in the ammonium and one in the nitride.

06:36 So there's two nitrogen on the reactant side and two nitrogen.

06:39 On the product side.

06:41 We have four hydrogens on the reactant side.

06:45 So just like in the previous problem, i'm going to have to put a two in front of h2o to give me four hydrogens on the product side.

06:52 Now let's check oxygens.

06:54 There are two oxygens on the reactant side, and two water molecules means that i have two oxygen on the product side.

07:02 So letter d is complete.

07:07 Let's go on to letter e.

07:09 Letter e, we have sodium hydrogen carbonate or sodium bicarbonate.

07:17 Hco3 is known by either of those names.

07:21 And it is decomposing to form sodium carbonate and water and carbon dioxide.

07:39 Okay, let's balance.

07:40 Reactant side, there's one sodium.

07:42 Product side, we have two sodiums.

07:45 So i'm going to start off by putting a two in front of my reactant.

07:49 That gives me two sodiums.

07:52 That also gives me two hydrogens now on the reactant side.

07:55 Looking at the product side, looking at the water, i see two hydrogens in the water.

08:02 I also have two carbons.

08:06 Well, there's one carbon right here, and there's another carbon over here.

08:12 So we have two carbons on the product side.

08:15 We have three oxygen on the reactant side.

08:18 I'm sorry, times two for a total of six.

08:21 So six oxygen on the reactant side, we have three oxygen in the na2 co3, plus another oxygen in the h2o, plus two more oxygen, gives me a total of six oxygen.

08:38 So this is balanced now.

08:45 All righty, let's go on to letter f.

08:51 In letter f, we have p4, u10, and it's reacting with six water molecules.

09:04 Whoops, let me go rid of the six for a minute.

09:07 If we don't know that yet.

09:09 It's reacting with water.

09:11 Sorry, i already had it balanced.

09:13 All right, so it's reacting with water, and it is forming h3 -p -o -4.

09:27 All right, starting with the phosphorus.

09:35 On the reactant side, i have four pses.

09:37 On the product side, i only have one.

09:39 So i'm going to have to put a coefficient of four in front of h3 -p -o -4 to give me four phosphorus on both sides.

09:51 Let's go on to the hydrogen.

09:54 I want to save the oxygen for last because it is in both of the reactants.

09:59 So going to the hydrogen on the product side, i have four times three, which is 12.

10:05 So i'm going to have to put a coefficient of six over here in front of the water to give me six times two or 12 hydrogen.

10:13 Now i can look at the oxygen.

10:16 There's 10 oxygen in the p4010 and six more oxygen in the h2o, or from the h2o.

10:24 So that gives me 16 oxygen.

10:26 On the product side, i have a coefficient of four and then a subscript of four after the oxygen.

10:33 So i need to multiply those.

10:34 Four times four gives me 16 oxygen.

10:39 So this one is now balanced.

10:45 All right.

10:47 We're almost halfway there.

10:49 Let's go on to letter g.

10:52 Give me some more space here to write.

10:56 And letter g.

10:58 We have hcl, reacting with calcium carbonate, producing calcium chloride plus water, plus carbon dioxide.

11:22 All right.

11:22 Once again, we have oxygen and a couple of molecules there on the product side, so i'm definitely going to save oxygen for last.

11:30 Let's start out with hydrogen.

11:32 React inside, there's one.

11:33 Product side, in h2o, there are two.

11:37 I go ahead and put a two in front of hcl.

11:41 Now as i look at cl, there are two cl on the reactant side, and we have a cl2 on the product side, so that means two chlorine there as well.

11:50 Calcium, there's one calcium on the reactant side, and one calcium on the product side, so that is balanced.

11:58 Carbon, there's one carbon on the reactant side, one on the product side.

12:03 That is balanced.

12:04 Let's check the oxygen out.

12:06 There's three oxygen on the reactant side.

12:09 And there's one oxygen in the water and two more oxygen in the carbon dioxide for a total of three.

12:15 So this is balanced just by adding that one coefficient in front of hcl.

12:23 Letter h.

12:27 In letter h, we have al reacting with h2 -s -o -4, sulfuric acid.

12:44 And it says that this makes or produces al2, so4, 3.

12:55 Plus h2.

13:00 Okay, let's go ahead and start balancing, starting with aluminum.

13:03 I'm just starting with the first element i come to, which is aluminum.

13:05 There's one on the reactant side, two on the product side.

13:09 So i need a two there...

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