Answer
The balanced equation for this reaction is:
$4NO_2(g) + 2H_2O(l) + O_2(g) -- \gt 4HNO_3(aq)$
Work Step by Step
1. Identify the chemical formula of each reactant and product:
Gaseous nitrogen dioxide: $NO_2(g)$
Liquid water: $H_2O(l)$
Gaseous oxygen: $O_2(g)$
Aqueous nitric acid: $HNO_3(aq)$
2. Write the equation:
$NO_2(g) + H_2O(l) + O_2(g) -- \gt HNO_3(aq)$
3. Balance the equation:
$N$: Products: 1; Reactants: 1
$O$: Products: 3; Reactants: 5
$H$: Products: 1; Reactants: 2
- Balance the number of hydrogens, by multiplying the coefficient of $HNO_3$ by 2:
$NO_2(g) + H_2O(l) + O_2(g) -- \gt 2HNO_3(aq)$
- Balance the nitrogen:
$2NO_2(g) + H_2O(l) + O_2(g) -- \gt 2HNO_3(aq)$
$N$: Products: 2; Reactants: 2
$O$: Products: 6; Reactants: 7
$H$: Products: 2; Reactants: 2
To remove an oxygen from the reactants side, we can multiply the coefficient of $O_2$ by $\frac{1}{2}$.
$2NO_2(g) + H_2O(l) + \frac{1}{2}O_2(g) -- \gt 2HNO_3(aq)$
Now, multiply all the coefficients by 2, to remove the fraction:
$4NO_2(g) + 2H_2O(l) + O_2(g) -- \gt 4HNO_3(aq)$
