Multiple Choice 1 point How many moles of oxygen gas (O_(2)) is needed to produce 18.0 liters of CO_(2) at STP? CH_(4(g))+2O_(2(g))arrow CO_(2(g))+2H_(2)O_((l))

## Step 1: <br />Identify the balanced chemical equation for the reaction. The given equation is already balanced:<br />\[ CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(l)} \]<br /><br />## Step 2:<br />Determine the mole ratio between \( CO_{2} \) and \( O_{2} \) from the balanced equation. According to the equation, 1 mole of \( CO_{2} \) is produced from 2 moles of \( O_{2} \).<br /><br />## Step 3:<br />Use the ideal gas law at STP (Standard Temperature and Pressure) to find the number of moles of \( CO_{2} \). At STP, 1 mole of any gas occupies 22.4 liters. Therefore, 18.0 liters of \( CO_{2} \) corresponds to:<br />\[ \frac{18.0 \text{ liters}}{22.4 \text{ liters/mole}} = 0.804 \text{ moles of } CO_{2} \]<br /><br />## Step 4:<br />Calculate the moles of \( O_{2} \) needed using the mole ratio from Step 2. Since 1 mole of \( CO_{2} \) requires 2 moles of \( O_{2} \), 0.804 moles of \( CO_{2} \) will require:<br />\[ 0.804 \text{ moles of } CO_{2} \times 2 \text{ moles of } O_{2} \text{ per mole of } CO_{2} = 1.608 \text{ moles of } O_{2} \]