How m For the following reaction: 2F_(2(g))+C_(2)H_(2(g))arrow C_(2)H_(2)F_(4(l)) How many liters of fluorine will be needed to make 45 grams of C_(2)H_(2)F_(4) (molar mass C_(2)H_(2)F_(4)=102.0298g/mol)

To determine how many liters of fluorine (\(F_2\)) are needed to make 45 grams of \(C_2H_2F_4\), we need to follow these steps:<br /><br />1. **Calculate the moles of \(C_2H_2F_4\):**<br /><br /> Given:<br /> - Mass of \(C_2H_2F_4\) = 45 grams<br /> - Molar mass of \(C_2H_2F_4\) = 102.0298 g/mol<br /><br /> Moles of \(C_2H_2F_4\) = \(\frac{\text{Mass}}{\text{Molar mass}}\)<br /><br /> \[<br /> \text{Moles of } C_2H_2F_4 = \frac{45 \text{ g}}{102.0298 \text{ g/mol}} \approx 0.441 \text{ moles}<br /> \]<br /><br />2. **Determine the stoichiometry of the reaction:**<br /><br /> The balanced chemical equation is:<br /> \[<br /> 2F_{2(g)} + C_{2}H_{2(g)} \rightarrow C_{2}H_{2}F_{4(l)}<br /> \]<br /><br /> From the equation, 1 mole of \(C_2H_2F_4\) is produced from 2 moles of \(F_2\).<br /><br />3. **Calculate the moles of \(F_2\) needed:**<br /><br /> Since 1 mole of \(C_2H_2F_4\) requires 2 moles of \(F_2\):<br /><br /> \[<br /> \text{Moles of } F_2 = 0.441 \text{ moles of } C_2H_2F_4 \times 2 = 0.882 \text{ moles of } F_2<br /> \]<br /><br />4. **Convert moles of \(F_2\) to liters:**<br /><br /> At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.<br /><br /> \[<br /> \text{Volume of } F_2 = 0.882 \text{ moles} \times 22.4 \text{ L/mol} \approx 19.75 \text{ liters}<br /> \]<br /><br />Therefore, approximately 19.75 liters of fluorine (\(F_2\)) are needed to make 45 grams of \(C_2H_2F_4\).