# Homework 39


Question 8.35

The unbalanced combustion equation for propane is:

$C_{3}H_{8}(g) + O_{2}(g) \rightarrow CO_{2}(g) + H_{2}O(g)$

How many moles of water vapor will be produced when 4.65 moles of propane reacts with excess oxygen gas?

Solution:

Step 1: Since the given equation is unbalanced, we must begin the solution by balancing the equation. Doing this will result in a balanced equation of:

$C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(g)$

Step 2: The question states that oxygen gas is in excess, so we do not need to worry about finding our limiting reagent. Thus, we can find the ratio of water vapor to propane. This ratio, according the balanced reaction, is:

$\frac{4mol\ H_{2}O(g)}{1mol\ C_{3}H_{8}(g)}$

Step 3: Using the ratio we found in Step 2, we can use dimensional analysis to find the moles of water vapor.

$4.65mol\ C_{3}H_{8}(g)\cdot \frac{4mol\ H_{2}O(g)}{1mol\ C_{3}H_{8}(g)}$

After cancelling the propane:

$4.65mol\ \cdot \frac{4mol\ H_{2}O(g)}{1mol\ }$

Step 5:

We are thus left with a simple multiplication problem where we can multiply 4.65 and 4.

After this, we find the number of moles of water vapor as:

${18.6mol\ H_{2}O(g)}$

$\mathit{18.6mol\ H_{2}O}$