Answer
The balanced equation for this reaction is:
$2SO_2(g) + O_2(g) + 2H_2O(l) -- \gt 2H_2SO_4(aq)$
Work Step by Step
1. Identify the chemical formula of each reactant and product:
Gaseous sulfur dioxide: $SO_2(g)$
Gaseous oxygen: $O_2(g)$
Liquid water: $H_2O(l)$
Aqueous sulfuric acid : $H_2SO_4(aq)$
2. Write the equation:
$SO_2(g) + O_2(g) + H_2O(l) -- \gt H_2SO_4(aq)$
3. Balance the equation:
$S:$ Products: 1; Reactants: 1;
$H:$ Products: 2; Reactants: 2;
$O:$ Products: 4; Reactants: 5: Unbalanced:
Since we have a compound like $O_2$, that only has oxygen in its formula, we can change its coefficient to balance the equation.
- We need to remove one oxygen from the reactants, so we can put a $\frac{1}{2}$ as the coefficient of $O_2$:
$SO_2(g) + \frac{1}{2}O_2(g) + H_2O(l) -- \gt H_2SO_4(aq)$
Now, to remove the fraction, multiply all the coefficients by 2:
$2SO_2(g) + 2*\frac{1}{2}O_2(g) + 2H_2O(l) -- \gt 2H_2SO_4(aq)$
$2SO_2(g) + O_2(g) + 2H_2O(l) -- \gt 2H_2SO_4(aq)$
