Answer
The balanced equation for this reaction is:
$2NO_2(g) + 7H_2(g) -- \gt 2NH_3(g) + 4H_2O(l)$
Work Step by Step
1. Identify the chemical formula of each reactant and product:
Gaseous nitrogen dioxide: $NO_2(g)$
Hydrogen gas: $H_2(g)$
Gaseous ammonia: $NH_3(g)$
Liquid water: $H_2O(l)$
2. Write the equation:
$NO_2(g) + H_2(g) -- \gt NH_3(g) + H_2O(l)$
3. Balance the equation:
$N$: Products: 1; Reactants: 1
$O$: Products: 1; Reactants: 2
$H$: Products: 5; Reactants: 2
- Balance the oxygen, because it only appears one time in each side:
$NO_2(g) + H_2(g) -- \gt NH_3(g) + 2H_2O(l)$
$H$: Products: 7; Reactants: 2;
Since the hydrogen only appears one time on the reactants side, we can multiply the $H_2(g)$ coefficient by $\frac{7}{2}$:
$NO_2(g) + \frac{7}{2}H_2(g) -- \gt NH_3(g) + 2H_2O(l)$
And multiply all the coefficients by 2:
$2NO_2(g) + 2*\frac{7}{2}H_2(g) -- \gt 2NH_3(g) + 2*2H_2O(l)$
$2NO_2(g) + 7H_2(g) -- \gt 2NH_3(g) + 4H_2O(l)$
