# Question e888d

Jan 17, 2015

Again, you don't need the balanced chemical equation for this one, all you need to know is the conservation of mass law.

In any chemical reaction, the total mass of the reactants must be equal to the total mass of the products. So, if you have $\text{21 kg}$ of gasoline and $\text{85 kg}$ on oxygen on the reactants' side, and assuming that you let all react, the total mass of the products must be equal to

$\text{21 kg + 85 kg = 106 kg}$

This is as straightforward as it gets. If however you need to determine how much of each product is produced, then the balanced equation is a must.

Usually, gasoline is expressed as octane in combustion reactions, which means that the balanced equation for its combustion is

$2 {C}_{8} {H}_{18} + 25 {O}_{2} \to 16 C {O}_{2} + 18 {H}_{2} O$

But, like I've said, for the combined mass of the products all you really need to know is the combined mass of the reactants.

Let's say you want to know exactly how much of each compound is formed. In this case, you must check for a limiting reagent. The number of moles of octane is

$21 \cdot {10}^{3}$ "g" * ("1 mole octane")/("114.0 g") = 184.2 $\text{moles}$

The number of moles of oxygen is

$85 \cdot {10}^{3}$ "g" * ("1 mole O"_2)/("32.0 g") = 2656# $\text{moles}$

According to the mole ratio between these two, each 2 moles of octane need 25 moles of oxygen, so

${n}_{\text{oxygen that reacts") = n_("octane}} \cdot \frac{25}{2} = 2303$ $\text{moles}$, which means that octane is the limiting reagent. The actual mass of oxygen that reacts is

$\text{2303 moles" * ("32.0 g")/("1 mole") = "73.7 kg}$

This means that the combined mass of the produced water and carbond dioxide is actually

$\text{21 kg" + "73.7 kg" = "94.7 kg}$

$\text{85 kg" - "73.7 kg" = "11.3 kg}$ of oxygen will not react. If you want to count this oxygen as well, the total combined mass of the products will again be $\text{106 kg}$.