Question

Chlorine and fluorine react to form gaseous chlorine trifluoride. Initially, 1.75 mol of chlorine and 3.68 mol of fluorine are combined. (Assume \(100 \%\) yield for the reaction.) (a) Write a balanced equation for the reaction. (b) What is the limiting reactant? (c) What is the theoretical yield of chlorine trifluoride in moles? (d) How many moles of excess reactant remain after reaction is complete.

Short answer

Answer: The limiting reactant is fluorine (F₂), the theoretical yield of chlorine trifluoride (ClF₃) is 2.453 moles, and the remaining moles of excess reactant (Cl₂) is 0.523 moles.

Step-by-step solution

Write a balanced equation for the reaction

For the given reaction, chlorine and fluorine react to form chlorine trifluoride. The unbalanced equation is: \(Cl_2 + F_2 \rightarrow ClF_3\) Balance the equation by adjusting the coefficients: \(Cl_2 + 3F_2 \rightarrow 2ClF_3\) The balanced equation for the reaction is: \(Cl_2 + 3F_2 \rightarrow 2ClF_3\)

Calculate the mole ratios of chlorine and fluorine

We have 1.75 moles of chlorine (\(Cl_2\)) and 3.68 moles of fluorine (\(F_2\)). According to the balanced equation, the mole ratio of \(Cl_2:F_2\) should be 1:3. Let's find the actual mole ratios: Actual mole ratio of \(Cl_2\) to \(F_2\) = \(\frac{1.75}{3.68}\)

Determine the limiting reactant

Based on the actual mole ratio calculated in step 2, compare it to the stoichiometric mole ratio (1:3) from the balanced equation to determine the limiting reactant. Actual mole ratio = \(\frac{1.75}{3.68} = 0.476\) Stoichiometric mole ratio = \(\frac{1}{3} = 0.333\) Since the actual mole ratio is greater than the stoichiometric mole ratio, \(F_2\) is the limiting reactant.

Calculate the theoretical yield of chlorine trifluoride

Now that we know the limiting reactant is \(F_2\), we can find the theoretical yield of \(ClF_3\). According to the balanced equation, 3 moles of \(F_2\) react with \(Cl_2\) to produce 2 moles of \(ClF_3\): Theoretical yield (moles of \(ClF_3\)) = \(\frac{2}{3} \times 3.68\ (moles\ of\ F_2) = 2.453\ mol\)

Calculate the remaining moles of the excess reactant

Since \(F_2\) is the limiting reactant, all of it will be consumed in the reaction. To find the remaining moles of \(Cl_2\), we need to calculate how much was used in the reaction: Moles of \(Cl_2\) reacted = \(\frac{1}{3} \times 3.68\ (moles\ of\ F_2) = 1.227\ mol\) Remaining moles of \(Cl_2\) = Initial moles of \(Cl_2\) - Moles of \(Cl_2\) reacted Remaining moles of \(Cl_2\)= \( 1.75 - 1.227 = 0.523\ mol\) The moles of excess reactant remaining after the reaction is complete are 0.523 mol of \(Cl_2\).