Question

Balance the following chemical equations.

Answer 1

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

$BaCl_2~+~H_2SO_4~=>~BaSO_4~+~2HCl$

We have 1 Ba, 2 Cl, 2 H and 4 O on both sides

Reaction 2

$Ca(OH)_2~+~CO_2~=>~CaCO_3~+~H_2O$

We have 1 Ca, 2O, 2 H and 1 C on both sides

Reaction 3

$2Al~+~3CuCl_2~=>~2 AlCl_3~+~3Cu$

We have 2 Al, 3 Cu, and 3 Cl on both sides

Reaction 4

$2SO_2~+~O_2~=>~2SO_3$

We have 2 A and 6O on both sides

Reaction 5

$2NH_3~+~3CuO~=>~3Cu~+~N_2~+~3H_2O$

We have 2 N, 3 H, 3 Cu and 3O on both sides.

I hope it helps