Question

1. chlorine gas reacts with fluorine gas to form chlorine trifluoride. \\(\ce{cl2(g) + 3f2(g) -> 2clf3(g)}\\) a 2.00-l reaction vessel, initially at 298 k, contains chlorine gas at a partial pressure of 337 mmhg and fluorine gas at a partial pressure of 729 mmhg. identify the limiting reactant and determine the theoretical yield of \\(\ce{clf3}\\) in grams. missed this? read section 6.7; watch iwe 6.12

Explanation:

Step1: Calculate moles of \( \text{Cl}_2 \) and \( \text{F}_2 \) using ideal gas law \( PV = nRT \)

For \( \text{Cl}_2 \): \( P = 337 \, \text{mmHg} = \frac{337}{760} \, \text{atm} \), \( V = 2.00 \, \text{L} \), \( T = 298 \, \text{K} \), \( R = 0.0821 \, \text{L·atm/(mol·K)} \)
\( n_{\text{Cl}_2} = \frac{P_{\text{Cl}_2}V}{RT} = \frac{\frac{337}{760} \times 2.00}{0.0821 \times 298} \approx 0.0363 \, \text{mol} \)

For \( \text{F}_2 \): \( P = 729 \, \text{mmHg} = \frac{729}{760} \, \text{atm} \)
\( n_{\text{F}_2} = \frac{P_{\text{F}_2}V}{RT} = \frac{\frac{729}{760} \times 2.00}{0.0821 \times 298} \approx 0.0783 \, \text{mol} \)

Step2: Determine limiting reactant using stoichiometry

From reaction \( \text{Cl}_2 + 3\text{F}_2
ightarrow 2\text{ClF}_3 \), mole ratio \( \text{Cl}_2:\text{F}_2 = 1:3 \)
Moles of \( \text{F}_2 \) required for \( \text{Cl}_2 \): \( 0.0363 \times 3 = 0.1089 \, \text{mol} \), but we have \( 0.0783 \, \text{mol} \) of \( \text{F}_2 \). So \( \text{F}_2 \) is limiting.

Step3: Calculate moles of \( \text{ClF}_3 \) from limiting reactant

Mole ratio \( \text{F}_2:\text{ClF}_3 = 3:2 \)
\( n_{\text{ClF}_3} = \frac{2}{3} \times n_{\text{F}_2} = \frac{2}{3} \times 0.0783 \approx 0.0522 \, \text{mol} \)

Step4: Calculate mass of \( \text{ClF}_3 \) (molar mass = 92.45 g/mol)

\( \text{Mass} = n \times \text{molar mass} = 0.0522 \times 92.45 \approx 4.82 \, \text{g} \)

Answer:

The theoretical yield of \( \text{ClF}_3 \) is approximately \( \boldsymbol{4.82 \, \text{g}} \) (limiting reactant is \( \text{F}_2 \))