How do you write an equation for the reaction between sodium chloride, sulfur dioxide gas, steam, and oxygen to give sodium sulfate and hydrogen chloride gas?

Oct 5, 2016

$4 N a C {l}_{\left(s\right)} + S {O}_{2 \left(g\right)} + 2 {H}_{2} {O}_{\left(l\right)} + 2 {O}_{2 \left(g\right)} \to 2 N {a}_{2} S {O}_{4 \left(a q\right)} + 4 H C {l}_{\left(g\right)}$

Explanation:

When you balance, start of with the most complicated molecules, and leave the simpler molecules to last. In this case, we want to balance $N {a}_{2} S {O}_{4 \left(a q\right)} , N a C {l}_{\left(s\right)} \mathmr{and} H C {l}_{\left(g\right)}$ first.

$2 N a C {l}_{\left(s\right)} + S {O}_{2 \left(g\right)} + {H}_{2} {O}_{\left(l\right)} + {O}_{2 \left(g\right)} \to N {a}_{2} S {O}_{4 \left(a q\right)} + 2 H C {l}_{\left(g\right)}$

We need two sodiums on the reactants side, so we place a two in front of sodium chloride. Then we need two chlorines as we just added two to sodium chloride, so we place a two in front of hydrogen chloride.

Now count your elements and see what's missing. We have 1 sulfur on both sides, 2 hydrogens, but we have 5 oxygens on the reactants side and only 4 on the products. Since the number of oxygens on the products can only be even, as it is in $S {O}_{4}$, we have to make the number of oxygens on the reactants even and then rebalance. The molecule that's making the number of oxygens odd is water, so we add a 2 to make it even and rebalance everything, getting:

$4 N a C {l}_{\left(s\right)} + S {O}_{2 \left(g\right)} + 2 {H}_{2} {O}_{\left(l\right)} + 2 {O}_{2 \left(g\right)} \to 2 N {a}_{2} S {O}_{4 \left(a q\right)} + 4 H C {l}_{\left(g\right)}$