Answer
The balanced equation for this reaction is:
$2NH_3(g) + CO_2(g) -- \gt CO(NH_2)_2(s) + H_2O(l)$
Work Step by Step
$NH_3(g) + CO_2(g) -- \gt CO(NH_2)_2(s) + H_2O(l)$
1. Start by balancing the elements that only appear once in each side:
Carbon:
We got 1 carbon on the reactants side, and 1 carbon on the products side. So it is already balanced:
$NH_3(g) + CO_2(g) -- \gt CO(NH_2)_2(s) + H_2O(l)$
Nitrogen:
We got 1 nitrogen on the reactants side, and 2 nitrogens on the products side. To balance we can multiply the nitrogen compound on the reactants by $2$:
$2NH_3(g) + CO_2(g) -- \gt CO(NH_2)_2(s) + H_2O(l)$
2. Now, balance the remaining elements:
$Oxygen:$
Products: 1 + 1 = 2 oxygens.
Reactants: 2*1 = 2 oxygens.
Balanced
$Hydrogen:$
Products: 2*2 + 2 = 6 hydrogens.
Reactants: 2*3 = 6 hydrogens.
The equation is balanced.
$2NH_3(g) + CO_2(g) -- \gt CO(NH_2)_2(s) + H_2O(l)$
