Balancing equations

- pooja195

Balancing equations

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- pooja195

\[\bf~\rm~CaCl_2+Al_2(C_2O_4)_3\rightarrow CaC_2O_4+AlCl_3\]

- bubblegum.

:)
1st did you try doing this?

- pooja195

No because i am hopelessly lost and i have no idea how to do this

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## More answers

- bubblegum.

oh okay :)
when we balance a chemical equation we just see that total number of atoms of 1 element are same on both sides of the equation
now the question is how do we find out total number of atoms of an element in one compound.
so to do this just follow these steps->|dw:1450035545688:dw|
now 1st your work is to get this point^

- pooja195

im confused :/

- bubblegum.

okay
do you know why a reaction is called unbalanced or balanced

- pooja195

nope ._.

- bubblegum.

okay
\(\bf~\rm~CaCl_2+Al_2(C_2O_4)_3\rightarrow CaC_2O_4+AlCl_3\)
in this equation can you find number of atoms of \(Ca~and~Cl\) in 1 molecule of \(CaCl_2\) which is on the left

- pooja195

Ca 1
Cl 2 ?

- bubblegum.

yea correct!
and now can you count the number of atoms of \(Ca~and~Cl\) on the right side

- pooja195

Ca 1
C1 3

- bubblegum.

yes correct :)
now compare the number of atoms of these 2 elements on the right and left
do u think there is something wrong it in :)

- pooja195

Yes

- pooja195

C1 2 <---right side
Cl 3 <---left side

- bubblegum.

yes :D
this is reason why the reaction is unbalanced!
and we are gonna balance it by multiplying the terms with suitable numbers
for example-
\(\color{red}{2}H_2+O_2=2H_2O\)

- pooja195

I see hmm
alright ok
soo
Can't we multiply
the 3 by 2
and 2 by 3 ?

- bubblegum.

\(\bf~\rm~\color{red}{3}CaCl_2+Al_2(C_2O_4)_3\rightarrow CaC_2O_4+\color{red}{2}AlCl_3\)
if you do this then the \(Ca\) atoms change on left
but the \(Al\) atoms gets balanced
now you gotta make evrything/every element balanced

- pooja195

See this is the part thats confusing >.<

- bubblegum.

ok so till now what can you say :)
which all elements are balanced?
you are close to the correct answer tho :D

- pooja195

\[\bf~\rm~2CaCl_2+1Al_2(C_2O_4)_3\rightarrow 2CaC_2O_4+1AlCl_3\]

- bubblegum.

no no
this is wrong
the \(Cl,~Al,~C\) atoms are unbalanced

- pooja195

;-;

- bubblegum.

\(\bf~\rm~\color{red}{3}CaCl_2+Al_2(C_2O_4)_3\rightarrow CaC_2O_4+\color{red}{2}AlCl_3\)
we had this^
now look at it carefully e_e
we see that we have balanced \(Cl,~Al\)
and now we have
3 atoms of \(Ca\) on left but only 1 on right
6 atoms of \(C\) on left but only 3 on right
12atoms of \(O\) on left but only 4 on right
now think to multiply something to \(CaC_2O_4\) :)

- pooja195

4 ;~;?

- bubblegum.

how did you get 4?

- pooja195

Wait would i distribute the 3?

- bubblegum.

sorry that will be-> 6 atoms of \(C\) on left and 2 on the right
yeah 3 is that number :)
just see if this fixes it all :) \(\color{red}{3}CaC_2O_4\)

- pooja195

;-; this is so confusing

- bubblegum.

do you agree that \(Cl~and~Al\) are balanced?

- pooja195

Yes

- bubblegum.

okay so
so far we have this-\(\bf~\rm~\color{red}{3}CaCl_2+Al_2(C_2O_4)_3\rightarrow CaC_2O_4+\color{red}{2}AlCl_3\)
Now
do you agree that ->
3 atoms of Ca on left but only 1 on right
6 atoms of C on left but only 2 on right
12atoms of O on left but only 4 on right

- pooja195

Yes

- bubblegum.

3 atoms of Ca on left but only 1 on right
6 atoms of C on left but only 2 on right
12atoms of O on left but only 4 on right
so what shall you multiply on the right^ to get what is on the left :)

- pooja195

1) 3 on the right
2) 3 on the right
3) 3 on the right

- bubblegum.

yeah! :D so we gotta multiply the right with a 3
and if we do this then \(C,~O~and~Ca\) get balanced and \(Al~and~Cl\) are already balanced
so can you rewrite the final equation now

- pooja195

\[\bf~\rm~3CaCl_2+3Al_2(C_2O_4)_3\rightarrow 3CaC_2O_4+3AlCl_3\]

- bubblegum.

jst a sec ima reload :)

- pooja195

o.O

- pooja195

buble gum come back T_T

- bubblegum.

m here typing :)

- bubblegum.

:) you don't have to multiply the whole equation
you only have to mutiply the part that you wanna fix
try to find out if what u did was correct or not by calculating if the number of atoms are balanced
you did till here->\(\bf~\rm~\color{red}{3}CaCl_2+Al_2(C_2O_4)_3\rightarrow CaC_2O_4+\color{red}{2}AlCl_3\)
and then u see that the unbalanced elements \(C,~O,~Ca\) on the left are thrice of that on the right
so you have to multiply by 3 on the right
\(\bf~\rm~\color{red}{3}CaCl_2+Al_2(C_2O_4)_3\rightarrow \color{blue}{3}CaC_2O_4+\color{red}{2}AlCl_3\)

- pooja195

OH ok makes a bit more sense THANK YOU SO MUCH <#3333

- bubblegum.

hhaaha np (;

- pooja195

;)

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