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

First tally the atoms. Since `SO_4^"2-"` is an ion, I'm going to consider it as one "atom" in order to not confuse myself.

`CaSO_4` + `AlCl_3` `rarr` `CaCl_2` + `Al_2(SO_4)_3` (unbalanced)

Based on the subscripts,

left side:
`Ca` = 1
(`SO_4`) = 1
`Al` = 1
`Cl` = 3

right side:
`Ca` = 1
(`SO_4`) = 3
`Al` = 2
`Cl` = 2

`color (red) 3CaSO_4` + `AlCl_3` `rarr` `CaCl_2` + `Al_2(SO_4)_3`

Let's start balancing the most complicated 'atom', `SO_4^"2-"`

left side:
`Ca` = (1 x `color (red) 3`) = 3
(`SO_4`) = (1 x `color (red) 3`) = 3
`Al` = 1
`Cl` = 3

right side:
`Ca` = 1
(`SO_4`) = 3
`Al` = 2
`Cl` = 2

Since `CaSO_4` is a substance, we need to also multiply the coefficient with its `Ca` atom. Now that there are 3 `Ca` atoms on the left, there must be 3 `Ca` atoms on the right.

`3CaSO_4` + `AlCl_3` `rarr` `color (blue) 3CaCl_2` + `Al_2(SO_4)_3`

left side:
`Ca` = (1 x 3) = 3
(`SO_4`) = (1 x 3) = 3
`Al` = 1
`Cl` = 3

right side:
`Ca` = (1 x `color (blue) 3`) = 3
(`SO_4`) = 3
`Al` = 2
`Cl` = (2 x `color (blue) 3`) = 6

Again notice that since `CaCl_2` is a substance, the coefficient 3 should also be applied to its `Cl` atoms. Since there are 6 atoms of `Cl` on the right, we need to also have the same number of `Cl` atoms on the left.

`3CaSO_4` + `color (green) 2AlCl_3` `rarr` `3CaCl_2` + `Al_2(SO_4)_3`

left side:
`Ca` = (1 x 3) = 3
(`SO_4`) = (1 x 3) = 3
`Al` = (1 x `color (green) 2`) = 2
`Cl` = (3 x `color (green) 2`) = 6

right side:
`Ca` = (1 x 3) = 3
(`SO_4`) = 3
`Al` = 2
`Cl` = (2 x 3) = 6

Again, since `AlCl_3` is a substance, the coefficient should also apply to the bonded `Al` atom.

Now the equation is balanced.