Answer
The balanced equation for this reaction is:
$C_3H_8(g) + 5O_2(g) --\gt 3CO_2(g) + 4H_2O(g)$
Work Step by Step
$C_3H_8(g) + O_2(g) --\gt CO_2(g) + H_2O(g)$
1. Start by balancing the elements that only appear once in each side:
Carbon:
We got 3 carbons on the reactants side, and 1 carbon on the products side. So we should multiply the carbon compound on the products by 3:
$C_3H_8(g) + O_2(g) --\gt 3CO_2(g) + H_2O(g)$
Hydrogen:
We got 8 hydrogens on the reactants side, and 2 hydrogens on the products side. To balance we can multiply the hydrogen compound on the products by $4$:
$C_3H_8(g) + O_2(g) --\gt 3CO_2(g) + 4H_2O(g)$
2. Now, balance the last element: $Oxygen:$
Products: 3*2 + 4*1 = 10 oxygens.
Reactants: 2*1 = 2 oxygens.
So, we should put a "5" next to the $O_2$, in order to balance the equation:
$C_3H_8(g) + 5O_2(g) --\gt 3CO_2(g) + 4H_2O(g)$
