#Na_2CO_3(s) + color(red)(2)HCl(aq) -> color(red)(2)NaCl(aq) + H_2O(l) + CO_2(g)#
The reasoning that you should follow in order to balance this reaction easily is:
1- Look at the #Na#. You have #2# in #Na_2CO_3# and #1# in #NaCl# then you should multiply #NaCl# by #color(red)(2)#.
2- When multiplying #NaCl# by 2, we will have to #Cl# in the products side, therefore, we should multiply #HCl# by #color(red)(2)#.
3- Look at the carbon atom #C# there is one atom at each side. No need for any action.
4- Look at the hydrogen atom #H# there is 2 atoms at each side. No need for any action.
5- Look at the oxygen atom #O# there is 3 atoms in #Na_2CO_3# and there is 1 atom in #H_2O# and 2 atoms in #CO_2# which makes it a total of 3 atoms. No need for any action.
