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Problem 14

Balance the following equations:

$$

\begin{array}{l}{\text { (a) } \mathrm{Ca}_{3} \mathrm{P}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{PH}_{3}(g)} \\ {\text { (b) } \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)} \\ {\text { (c) } \mathrm{AgNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow \mathrm{Ag}_{2} \mathrm{CO}_{3}(s)+\mathrm{NaNO}_{3}(a q)} \\ {\text { (d) } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)}\end{array}

$$

Answer

a) See explanation for solution.

b) See Explanation

c) See Explanation

d) $4 C_{2} H_{5} N H_{2}+15 O_{2} \rightarrow 8 C O_{2}+14 H_{2} O+2 N_{2}$

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## Discussion

## Video Transcript

Okay, Sort of balance. The reaction equations. We will follow some simple rules. First, we will balance the metals. Dan will balance non metals and the poly atomic ions Eve. The poly atomic ions are intact in both side of variation. Then we will balance Hide Rosen's. And finally, we'LL balance oxygen. I can always started our first reaction here, which is sea. It repeats regulate its tool to form. See weighs two and peace train. Okay, so this right don't you are Ah, Antos. Orion present on its side. Calcium. It was for us. High treason and oxygen. And this side we have again till she, um for us for us. Hydrogen and oxygen. So you have already, right? No, within them in order of metal nonmetal hydrogen oxygen. The first real balance or metal in the reactor side. We have three calcium. Where I don't three. Let's first right down. Um, how many off each? Only him on this, if you have one tells him and then we have to force for us in the reactor. Inside. We have one for you. For us in the morning. We have to head Rosen in the reactor sign. Perhaps that we have to hide. Resent. Plus three total five high treason. And we have one oxygen, the reactor sign. And we have to look Sian in the black site. All right. Says if said, Well, first man is tomato. So since we have three calcium in the reactor inside, we need to get him in the product side. And that's why I will multiply this to see aways to wait three. Now we have, um, three kills him in the reactor inside also. Yeah, but since we have also market right oxygen, hydrogen, and we have to change these numbers, So now in this side, we have three times two, six oxygen and we have three times two six, plus three that men's nine. I draw them. Okay, now we got our gal. She imbalanced. Now we have to balance our force for us. Okay, so here we have two firsts for us, and here we have one. That means we need to force ready the product side. And for that we will place it to here, which will give us two firsts for us. And in the same day, we have also j the number of high treason. No, here two times three six plus three times two six, six extra will hide risen. Now we got our first for us Balanced. Now we have to bed. And so are you, Drizin. So we have grilled hydrogen peroxide, but two hundred in the reactor site, like mints. We need six more. Ah, we made ten more. I'd rather in the reactor sign. So if we just multiply this by six, we can have a little present in the reactors that also to give us systems to twelve in prison, in the reactors side. Also. Now we have our address and balanced nobody to another oxygen. So yes, and again says we're multiplied oxygen also bay six. So we have six socks. Is that in the return sign? And we are already have having six oxygen. The product side that Mr Rocks, it is already balanced. Tonight we check. We have trickled. Shimoni side were to force for us on its side. We have grilled. I'd return East side and six socks in his sight. That means our each atom or eye on presenting its side is balanced. And that means this is our balanced reaction in creation for destruction Now let's go to our second one. Here. Ah, spaces we have is aluminium, Dan. We have. So our non metal here is the poly atomic eye on which is self fat. Well, great it yourself it then we have ID Rosen and oxygen. And in this side with him aluminium so fat I draws them and oxygen. All right, now let's write don't How many of which we have. We have one aluminium here, one self it Here we have three plus two, five dead, risen. And we have, um, three oxygen. See, this arson is entering the self. It will not touch it. And since Alpha is intact in both Syria, writing in a self so we have a three other oxygen here in this side. We have two aluminium, three self at. I have to had risen and one oxygen. Now, let's start with our, um, aluminium here. So we have to eliminate Min the product side that Miss Renick dwelling within the reactor inside and that for that we just multiply this and with the right to which will give us two aluminium. But at the same time, we're changing the number of oxygen and hydrogen Now we have two times two six plus two eight had risen. We have eight had risen and we have two times two six oxygen. Okay, so we have menaced Are metal aluminium now will balance yourself it So you have three cell foot in them on the product saying that means we need three sulphur in the reactor site. Multiply this by three. That means what we'LL do is we have three solvent. But we're also changing the number of hydrogen that Mr Teams toe six had risen. Bless you three times. Two six. That means we have six plastics to avoid prison. Good. Okay, so now we have our cell fertile. So balanced, nor any Tobin is the head risen? So, in a reactor site, we have pulled by prison. When the peroxide we have only two. That means we need to multiply this age. Two of six tohave sex timeto twelve a treason. But we're also changing one hour of oxygen. It means we have ah, six oxygen Now. Now, if we look att the numbers of each pacific and say that we have two aluminium on its side, we have three suffered on his sign. We're to a hydrogen on its side and six oxygen on its side. And that means this is our balanced equation for this creation. Now let's go to our next reaction. Where our spaces there, silver. And since night Crate is injecting both side and also carbonic injecting. Both side will write them as nitrogen carbon et We have men on three. We have seen a tree and we have another metal sodium Celeste, right, sodium on the Tom. And here was the right sodium silver in or three. I see a tree. Okay, let's illustrate. Don't how many of this we have been inside. We have to sodium here and we have one sodium in the product. We have won silver in the reactant to silver in the product. We have one night treating the reactant with a one nighter in the product. We have one corroborating the reactant. We have one car burning the product. All right, so now first two bandits are sodium. We have to so dam in the reacting, but we have one. So I came in the product. That missile multiply this by to get to sodium in the protect. Maxie's were also multiplying nitrate that, miss. Now we have two nitrate in the product also. Now we've got our story in balance. Now we have to us or silver one still ran the reactor, but to sell When the product that's I will multiply this by two. You have two silver in the reactor and also at the same time. We're multiplying nitrate by. So that means we have two nitrate in the reactor. And also then, ah, we have toe. So now we can see that we have our sodium silver nitrate, all our penance and also our car murders also balanced that men's since all this Mrs Air balanced in both sides, This's our imbalanced chemical equation for destruction. Now we will go our final on here. What we have. Um so since we don't have any metal here, we start with carbon. Then we have night Rosen. Then we have had risen and oxygen and this side we have same carbon make prison one carbon, two natures and two hundred zin and two plus one three oxygen. Now let's start balancing your carbon. We have one carbon in the product, but to covering the reactor, so will multiply. They seal to buy to get two carbon in the product at the same time. We now have four oxygen in the product off four plus one five oxygen. Now we have two venison actress and we have two. Night was in the parade but one night within the reactor And that means we need to multiply Thies Bye Toto have to nitrous and in the reactor doll song. But at the same damn, we have changed the number ofthe carbon so we have four carbon in the reactor now and we have also changed the number of hydro's. And so we have. If I place to seven times tumors, Fortin had risen. All right, so that means we need to Jane Jar. Since we obtained a number of carbon atoms in the reactor inside, we need to change the car. Madam in the peroxide also says he had four hearing before carbon in the product side. Also that men's we'll change it to four tohave for carbon in the production line. But on the same thing we want to change the number of oxygen now have four times to eight must one nine oxygen in the production line and then now we have our carbon and nitrogen and balanced so far. Then we have to balance our hide. Risen, we have forty nitrogen in the reactors side. That means waited fourteen hydrogen in the product side also. And for that we just multiply this by seven tto gett fortin had risen the same thing. We're changing the number of oxygen Now we have seven plus eight that Mr Lian in the reactor site. Okay, so now we have our carbon, nitrogen and hydrogen balanced. But we have to balance our oxygen now. All right, so we have some oxygen in the reactors side and we have fifteen oxygen in the correct side. That man's we need fifteen oxen in the reactor inside also. So if we want fifteen oxy so seven point five time stool, give us fifteen. So since we cannot write a fraction in the a zit coefficient, you know, chemical equation What? Well, right. We'LL write it by as fifteen two So fifteen by two men seven point five times to Now we have fifteen oxy and also in the reactor site. So now we have our e question balanced. But with this way, we have to have a whole number here. And for that what we'LL do with simply multiplied the holy question by two to have a whole number here. So if we multiply the holy question by two, we'LL have two times two four secrets. Women is too. They want to play this by two. We'll have simply and then four time stool He was eight and then seven times too will give us fourteen. And finally, um, we have two in tow. So this is the balanced equation for this reaction.

## Recommended Questions

Balance the following equations:

$$

\begin{array}{l}{\text { (a) } \mathrm{Al}_{4} \mathrm{C}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{CH}_{4}(g)} \\ {\text { (b) } \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (c) } \mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)} \\ {\text { (d) } \mathrm{Mg}_{3} \mathrm{N}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{MgSO}_{4}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q)}\end{array}

$$

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$$

\begin{array}{l}{\text { (a) } \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)} \\ {\text { (b) } \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HNO}_{3}(a q)} \\ {\text { (c) } \mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(l)+\mathrm{HCl}(g)} \\ {\text { (d) } \mathrm{Zn}(\mathrm{OH})_{2}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)}\end{array}

$$

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$$

\begin{array}{l}{\text { (a) } \mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Li}_{3} \mathrm{N}(s)} \\ {\text { (b) } \mathrm{TiCl}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{TiO}_{2}(s)+\mathrm{HCl}(a q)} \\ {\text { (c) } \mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (d) } \mathrm{AlCl}_{3}(s)+\mathrm{Ca}_{3} \mathrm{N}_{2}(s) \longrightarrow \mathrm{AlN}(s)+\mathrm{CaCl}_{2}(s)}\end{array}

$$

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$$

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$$

\begin{array}{l}{\text { a. } \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q) \rightarrow} \\ {\mathrm{BaCrO}_{4}(s)+\mathrm{NaNO}_{3}(a q)} \\ {\text { b. } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)} \\ {\text { c. } \mathrm{CaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g)}\end{array}

$$

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\begin{array}{l}{\text { (a) } \mathrm{ZnCO}_{3}(s) \stackrel{\Delta}{\longrightarrow}} \\ {\text { (b) } \mathrm{BaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow} \\ {\text { (c) } \mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow}\end{array}

$$

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