5. How many liters of propane (C_(3)H_(8)) are needed to react with 15.2 lited of oxygen at STP in order to produce water and carbon dioxide? 1C_(3)H_(8)+5O_(2)arrow 4H_(2)O+3CO_(2)

## Step 1: <br />The balanced chemical equation for the reaction is given as:<br />\[1C_{3}H_{8}+5O_{2}\rightarrow 4H_{2}O+3CO_{2}\]<br />This equation tells us that 1 liter of propane reacts with 5 liters of oxygen to produce water and carbon dioxide.<br /><br />## Step 2:<br />We are given that 15.2 liters of oxygen are available. We need to find out how many liters of propane are needed to react with this amount of oxygen.<br /><br />## Step 3:<br />We can set up a proportion to solve for the unknown quantity of propane. The proportion is set up as follows:<br />\[\frac{1 \text{ liter of propane}}{5 \text{ liters of oxygen}} = \frac{x \text{ liters of propane}}{15.2 \text{ liters of oxygen}}\]<br /><br />## Step 4:<br />Cross-multiply and solve for \(x\):<br />\[5x = 15.2\]<br />\[x = \frac{15.2}{5}\]