Question 3.18: Using Molecular Depictions in a Limiting-Reactant Problem Pr......

Plan (a) We have to find the limiting reactant. The first step is to write the balanced equation, so we need the formulas and states of matter. From the name, chlorine trifluoride, we know the product consists of one Cl atom bonded to three F atoms, or ClF_3. Elemental chlorine and fluorine are the diatomic molecules Cl_2 and F_2, and all three substances are gases. To find the limiting reactant, we find the number of molecules of product that would form from the numbers of molecules of each reactant: whichever forms less product is the limiting reactant. (b) We use these numbers of molecules to write a reaction table. (c) We use the numbers in the Final line of the table (Table 1) to draw the scene.

Solution (a) The balanced equation is

                Cl_2(g)  +  3F_2(g)  ⟶  2ClF_3(g)
        \text{For }Cl_2\text{: Molecules of } ClF_3  =  3  \cancel{\text{molecules of }Cl_2}  \times  \frac{2  \text{molecules of }ClF_3}{1  \cancel{\text{molecules of }Cl_2}}
                           =  6 \text{ molecules of } ClF_3
        \text{For }F_2\text{: Molecules of } ClF_3  =  6  \cancel{\text{molecules of }F_2}  \times  \frac{2  \text{molecules of }ClF_3}{3  \cancel{\text{molecules of }F_2}}
                           =  4 \text{ molecules of } ClF_3

Because it forms less product, F_2 is the limiting reactant.

(b) Since F_2 is the limiting reactant, all of it (6 molecules) is used in the Change line of the reaction table (Table 1):

(c) The representative portion of the final reaction mixture includes 1 molecule of Cl_2 (the reactant in excess) and 4 molecules of product ClF_3.

Check The equation is balanced: reactants (2 Cl, 6 F) → products (2 Cl, 6 F). And, as shown in the circles (Fig 3.18b), the numbers of each type of atom before and after the reaction are equal. Let’s think through our choice of limiting reactant. From the equation, one Cl_2 needs three F_2 to form two ClF_3. Therefore, the three Cl_2 molecules in the circle depicting reactants need nine (3 × 3) F_2. But there are only six F_2, so there is not enough F_2 to react with the available Cl_2; or put the other way, there is too much Cl_2 to react with the available F_2. From either point of view, F_2 is the limiting reactant.

Comment Notice that there are fewer Cl_2 molecules than F_2 molecules initially. The limiting reactant is not the reactant present in the smaller amount, but the reactant that forms less product.