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.
