# How would you balance: CaSO4+AlCl3→CaCl2+Al2(SO4)3?

Nov 7, 2015

$3 C a S {O}_{4}$ + $2 A l C {l}_{3}$ $\rightarrow$ $3 C a C {l}_{2}$ + $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$

#### Explanation:

First tally the atoms. Since $S {O}_{4}^{\text{2-}}$ is an ion, I'm going to consider it as one "atom" in order to not confuse myself.

$C a S {O}_{4}$ + $A l C {l}_{3}$ $\rightarrow$ $C a C {l}_{2}$ + $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$ (unbalanced)

Based on the subscripts,

left side:
$C a$ = 1
($S {O}_{4}$) = 1
$A l$ = 1
$C l$ = 3

right side:
$C a$ = 1
($S {O}_{4}$) = 3
$A l$ = 2
$C l$ = 2

$\textcolor{red}{3} C a S {O}_{4}$ + $A l C {l}_{3}$ $\rightarrow$ $C a C {l}_{2}$ + $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$

Let's start balancing the most complicated 'atom', $S {O}_{4}^{\text{2-}}$

left side:
$C a$ = (1 x $\textcolor{red}{3}$) = 3
($S {O}_{4}$) = (1 x $\textcolor{red}{3}$) = 3
$A l$ = 1
$C l$ = 3

right side:
$C a$ = 1
($S {O}_{4}$) = 3
$A l$ = 2
$C l$ = 2

Since $C a S {O}_{4}$ is a substance, we need to also multiply the coefficient with its $C a$ atom. Now that there are 3 $C a$ atoms on the left, there must be 3 $C a$ atoms on the right.

$3 C a S {O}_{4}$ + $A l C {l}_{3}$ $\rightarrow$ $\textcolor{b l u e}{3} C a C {l}_{2}$ + $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$

left side:
$C a$ = (1 x 3) = 3
($S {O}_{4}$) = (1 x 3) = 3
$A l$ = 1
$C l$ = 3

right side:
$C a$ = (1 x $\textcolor{b l u e}{3}$) = 3
($S {O}_{4}$) = 3
$A l$ = 2
$C l$ = (2 x $\textcolor{b l u e}{3}$) = 6

Again notice that since $C a C {l}_{2}$ is a substance, the coefficient 3 should also be applied to its $C l$ atoms. Since there are 6 atoms of $C l$ on the right, we need to also have the same number of $C l$ atoms on the left.

$3 C a S {O}_{4}$ + $\textcolor{g r e e n}{2} A l C {l}_{3}$ $\rightarrow$ $3 C a C {l}_{2}$ + $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$

left side:
$C a$ = (1 x 3) = 3
($S {O}_{4}$) = (1 x 3) = 3
$A l$ = (1 x $\textcolor{g r e e n}{2}$) = 2
$C l$ = (3 x $\textcolor{g r e e n}{2}$) = 6

right side:
$C a$ = (1 x 3) = 3
($S {O}_{4}$) = 3
$A l$ = 2
$C l$ = (2 x 3) = 6

Again, since $A l C {l}_{3}$ is a substance, the coefficient should also apply to the bonded $A l$ atom.

Now the equation is balanced.