4. How many moles of CO will react with 0.6573 moles of 02? 4C_(2)H_(3)OCl+5O_(2)arrow 8CO+6H_(2)O+2Cl_(2)

## Step 1<br />The balanced chemical equation given is \(4C_{2}H_{3}OCl+5O_{2}\rightarrow 8CO+6H_{2}O+2Cl_{2}\). This equation tells us that 5 moles of \(O_{2}\) react with 4 moles of \(C_{2}H_{3}OCl\) to produce 8 moles of \(CO\).<br /><br />## Step 2<br />We are given that 0.6573 moles of \(O_{2}\) are present. We need to find out how many moles of \(CO\) will be produced.<br /><br />## Step 3<br />We can set up a proportion to solve for the unknown number of moles of \(CO\). The proportion is set up as follows:<br /><br />### \(\frac{5 \text{ moles of } O_{2}}{8 \text{ moles of } CO} = \frac{0.6573 \text{ moles of } O_{2}}{x \text{ moles of } CO} \)<br /><br />## Step 4<br />Solving for \(x\), we get:<br /><br />### \(x = \frac{0.6573 \times 8}{5} = 1.0512 \text{ moles of } CO \)