let's do some practice problems for

balancing chemical equations we'll start

off with some examples that are pretty

basic and straightforward and then the

problems will get more challenging as we

move on here's our first equation we got

xenon and fluorine we want to keep track

of how many atoms of these elements we

have on both sides of the equation so

we're going to make a little chart okay

I got xenon and fluorine on this side

and then xenon and fluorine on this side

over here I have one xenon atom one and

then I have f2 so I got two fluorines

over here also one xenon and then f6

so I got 6 fluorines now take a look at

these numbers this equation isn't

balanced yet because we have different

numbers of the atoms for one of the

elements i've got two fluorine here but

i've got six fluorine here so it's not

balanced in order to balance it I've got

to add numbers or coefficients in front

of one or more of these elements and

compounds to change the number of atoms

that I have on the different sides of

the equation okay here's how I'm going

to do I got six fluorine here but I got

two fluorine here so for this to balance

I need more fluorines on the left side I

can add a number in front of the f2 here

if I put 3 in front of this f2 here

we'll put it in I put 3 in front of this

f2 now I have three times two gives me

six fluorine and now they balance one

one for the Xenon's 6 and 6 for the

fluorine so now this is a balanced

equation by adding this number by adding

this coefficient now really quick this

is one very common question people often

ask wait why did you put that 3 there

couldn't you just change this 2 to a 6

and then it would also balance no no no

no no no no no no you can't do that

super common misconception you can't

change these subscripts here you can't

change them you can't add them so you

can't change this to a six

you can't put a six here or anything

like that the only thing you can do is

you can put numbers in front of the

elements or compounds okay but you can't

change or add the subscripts okay so

that's how you balance an equation let's

do a whole bunch more practice okay so

in this equation we've got three

elements silver AG hydrogen H and sulfur

s obviously we've got these on both

sides of the equation okay so let's see

how many atoms we have of each over here

we got one AG we got h2 so we got two

hydrogen atoms and we have s so we got

one sulfur atom over here we have AG two

so we have two silvers here on sulfur we

have one and then we have h2 so we have

two hydrogens okay what balances and

what doesn't well the hydrogen's and

sulfur both balance but the silver we

have two here and one here we can fix it

by adding a number or coefficient in

front of one or more of these elements

and compounds

I need more silvers on this side and

luckily I can just put a two in front of

this AG so now instead of having one AG

now I have two and now I got two two and

one it balances this equation is a

little bit more challenging because we

have oxygen in all the compounds here

that means we're just going to have to

be a little bit more careful when we're

adding up the number of oxygen atoms we

have okay so the elements in this

equation we got K potassium Oh oxygen H

hydrogen and C carbon and over on this

side we've got potassium oxygen hydrogen

and carbon okay how many of each of

these do we have okay we got one

potassium here now for oxygen this is

where you just got to be a little bit

more careful okay because we have one

oxygen here but then we have two

Oxygen's here so one plus two total we

have three then hydrogen we got one of

those and carbon we got one of those

over here for potassium we have two of

these because we got k2 then oxygen we

have three there and we have one there

it's going to give us four total then we

have h2 we have two hydrogen's and one

carbon okay so what balances what

doesn't the carbons balance but other

than that we've got to add some

coefficients to change this stuff around

okay oxygen doesn't balance but since

oxygen is in every one of the compounds

I'm going to leave that for later I'm

going to start with potassium because

it's easier right I have two potassium

here but I've only got one here so the

first thing I'm going to do is I'm going

to put a two in front of this Koh and

let's see what that does

okay the first thing that's going to do

is it's going to give me two instead of

one potassium because I have the two

times K now four oxygens let's look at

how it's going to change that now I'm

going to have two oxygens because two

times that plus the two that I have over

there in co2 so I'm going to have two

plus two is now going to give me four so

I've got four oxygens and in terms of

hydrogen now I'm going to have two times

one hydrants I'm going to get two

hydrogens and check it out putting this

putting this two here changed everything

so now I've got two four two one it all

balances now we're going to start doing

some equations that require more than

one step to balance okay so this one

here has sodium and chlorine over here

we got one sodium and we have CL two we

have two chlorine atoms over here we

have NaCl so we have one sodium one na

and one Cl okay how am I going to start

well I have more chlorine on this side I

have two over here and I only have one

here okay so I can start out by putting

it two in front of na

see oh okay that's going to be two CL

but look what else it's going to do okay

now I have two times n a so I have two n

A's I also have two times CL okay so I

also have two CLS so now the CLS

balanced but now I have two n A's to

sodium on this side and only have one on

this side so now to balance this out I'm

going to put it - in front of this na

here and now I have two of these so now

on both sides of the equation I have two

sodium's

and two chlorines the equation balances

okay

iron oxygen and carbon over here I have

one iron one oxygen and one carbon here

I have one iron two oxygens and one

carbon okay how am I going to balance

this the first thing I'm going to do is

I'm going to balance these oxygens okay

so I have two over here and I have one

here I'm going to change that by putting

a 2 in front of the f EO okay so that's

going to give me two oxygens but it's

also now going to multiply this Fe by 2

so now I'm going to have 2 Fe ok so now

I've got 2 Fe here and one there so now

these don't balance so the next thing

I'm going to do is I'm going to put a 2

in front of this Fe there so now instead

of having one I have two of them now I

have two two one

it balances ok let's look what we have

here we have silicon on this side we

have one of them o to two oxygen and one

carbon over here we have one silicon

oxygen we have one and carbon we have

one plus one so we have two ok so what

doesn't balance well the carbons don't

balance and the oxygens don't balance

I'm going to leave the carbon alone for

now because it's in both of these

compounds so I kind of don't want to

mess with it right now but I could I'm

just going to start out with the oxygen

okay I have two oxygens on this side and

I have one oxygen on this side so I'm

going to multiply the CE o by 2 so that

can give me two oxygens but now it's

also going to do something else right

because we're also multiplying this C by

that too so that means now for the total

carbons I'm gonna have the one from

there plus now the two from there and

that's going to give me a total now of

three okay so now I have three carbons

here I've got one carbon here so I'm

going to put a 3 in front of this carbon

that's the only carbon on this side so

now I have 3 times 1 is 3 1 2 and 3 and

they balance this equation has 5

different elements in it and it also has

these parenthesis so we'll talk about

these in a minute first the number of

atoms iron Fe we have one here CL 3 na 1

o and H we have one of those okay now Fe

over here we have 1 and then we got the

parenthesis okay so the parenthesis mean

that everything inside here is

multiplied by 3 okay so we have Oh times

3 which means we have 3 OS and we have H

times 3 so that's three H's and then we

have Na and Cl so one each of those ok

so where are we going to start with us

well I have I have an imbalance in my

oxygens here so I might as well just

start there okay so three here one over

here so I'm going to put a 3 in front of

NaOH okay so that is going to multiply

all the stuff by three I have three na 3

o + 3 H ok what does that do for me

well that balance the oxygens and

hydrogen's but now the sodium na I have

three on this side I don't have one on

this side okay so I can fix that by

multiplying this NaCl by three here okay

so that's going to give me three na and

it's also going to be three times a CL

so now I have three CL and check it out

that fixed it I have 3 CL over here but

I got 3 CL there great so 1 3 3 3 3 it

balances we got some more parentheses

here let's take a look at this equation

okay I have one aluminum two hydrogen's

one sulfur and four oxygens then over

here I have two aluminum's everything in

here is multiplied by three so that

means that I have three Sulphurs and I

have four times three which is 12

oxygens and then I have two hydrogen's

okay what am I going to do first well

take a look at this we have aluminum and

we have hydrogen on their own so I'm

going to save them for last I don't want

to balance them right now because we can

use these at the end to straighten out

some of these details

so let's look right now it's sulfur and

oxygen

I can fix both the sulfur and the oxygen

and balance them by multiplying these by

three right I have three sulfur's and

twelve oxygen so if I just multiply

these by three I can get them to balance

so I've got sulfur and oxygen there so

I'm going to start by putting a 3 in

front of that okay

that's then going to give me three

Sulphurs and three times four twelve

oxygens and it's also going to give me

some more hydrogens okay so now I got

six hydrogens okay so we got the sulfur

and the oxygen to balance now just by

multiplying by three now my hydrogen my

hydrogen doesn't balance I have six on

this side two on this side so I can fix

that by multiplying the hydrogen by

three now I have three times two on this

side which is going to give me six and

then take a look at the aluminum I have

two on this side one on this side but I

can fix that by multiplying by two over

here and then I get those to balance and

everything's good you may have noticed

these are getting a little bit more

complex it takes more and more steps to

balance them okay so for this we have

one nitrogen we have three hydrogens we

have one Cu which is copper and we have

one oxygen over here we have two

nitrogens

at two hydrogen's one copper and one

oxygen okay where am I going to begin

with us

well copper

nitrogen are on their own here so I'm

going to leave them to last I'm going to

focus in on hydrogen and oxygen though

the oxygens are balanced right now but

the hydrogen's are not and check out

I've got this 3-2 thing going on I can

solve that by doing a criss cross

multiplication to get 6 so the

hydrogen's are here I'm going to

multiply this by 2 and then I'm going to

multiply this by 3 ok what that's going

to do is it's going to give me 2 times 3

6 hydrogen's it's also going to give me

2 nitrogen's and then over here it's

going to give me 3 times 2 hydrogen's

which which is 6 and then 3 times 1

which is 3 oxygens ok so now I've got my

hydrogen's balanced the next thing that

I can do is um leaving copper and

nitrogen for later let's balance the

oxygens ok I have 3 on this side I have

one on this side my oxygen is there so

I'm going to multiply this by 3 so I'm

going to do 3 times 1 get 3 oxygens but

I'm also going to get 3 times 1 for

copper ok so now my oxygen is balanced

but my copper my copper is still a

problem here

ok so what I'm going to do there is I

need 3 on this side because I have 3 on

that side but luckily coppers by itself

so that's what I say like at the very

end you can go through and fix this

stuff so I'll put a 3 there in front of

copper so 3 times 1 gives me 3 for

copper so this balances now