Answer:
See the explanation
Explanation:
In this question we have to take into account that we have to get the same amount of atoms on both sides, so:
Reaction 1
[tex]BaCl_2~+~H_2SO_4~=>~BaSO_4~+~2HCl[/tex]
We have 1 Ba, 2 Cl, 2 H and 4 O on both sides
Reaction 2
[tex]Ca(OH)_2~+~CO_2~=>~CaCO_3~+~H_2O[/tex]
We have 1 Ca, 2O, 2 H and 1 C on both sides
Reaction 3
[tex]2Al~+~3CuCl_2~=>~2 AlCl_3~+~3Cu[/tex]
We have 2 Al, 3 Cu, and 3 Cl on both sides
Reaction 4
[tex]2SO_2~+~O_2~=>~2SO_3[/tex]
We have 2 A and 6O on both sides
Reaction 5
[tex]2NH_3~+~3CuO~=>~3Cu~+~N_2~+~3H_2O[/tex]
We have 2 N, 3 H, 3 Cu and 3O on both sides.
I hope it helps