Answer
The balanced equation for this reaction is:
$2NO(g) + 2CO(g) -- \gt N_2(g) + 2CO_2(g)$
Work Step by Step
$NO(g) + CO(g) -- \gt N_2(g) + CO_2(g)$
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:
$NO(g) + CO(g) -- \gt N_2(g) + CO_2(g)$
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 products by $\frac{1}{2}$:
** Notice: it is more adequate to change the coefficient of a compound with only one type of element $N$.
$NO(g) + CO(g) -- \gt \frac{1}{2}N_2(g) + CO_2(g)$
Now, multiply all the coefficients by 2, to remove the fraction:
$2NO(g) + 2CO(g) -- \gt N_2(g) + 2CO_2(g)$
2. Now, balance the last element: $Oxygen:$
Products: 2*2 = 4 oxygens.
Reactants: 2*1 + 2*1 = 4 oxygens.
The equation is already balanced.
