Question 3.19: Calculating Quantities in a Limiting-Reactant Problem: Amoun......
Calculating Quantities in a Limiting-Reactant Problem: Amount to Amount
Problem In another preparation of ClF_3 (see Sample Problem 3.18), 0.750 mol of Cl_2 reacts with 3.00 mol of F_2. (a) Find the limiting reactant. (b) Write a reaction table.
Plan (a) We find the limiting reactant by calculating the amount (mol) of ClF_3 formed from the amount (mol) of each reactant: the reactant that forms fewer moles of ClF_3 is limiting.
(b) We enter those values into the reaction table (Table 1).
Solution (a) Determining the limiting reactant:
Finding amount (mol) of ClF_3 from amount (mol) of Cl2:
\text{Amount (mol) of }ClF_3 = 0.750 \cancel{\text{mol} Cl_2} \times \frac{2 \text{mol} ClF_3}{1 \cancel{\text{mol} ClF_2}} = 1.50 \text{mol} ClF_3Finding amount (mol) of ClF_3 from amount (mol) of F_2:
\text{Amount (mol) of }ClF_3 = 3.00 \cancel{\text{mol} F_2} \times \frac{2 \text{mol} ClF_3}{3 \cancel{\text{mol} F_2}} = 2 \text{mol} ClF_3In this case, Cl_2 is limiting because it forms fewer moles of ClF_3.
(b) Writing the reaction table (Table 1), with Cl_2 limiting:
Check Let’s check that Cl_2 is the limiting reactant by assuming, for the moment, that F_2 is limiting. If that were true, all 3.00 mol of F_2 would react to form 2.00 mol of ClF_3. However, based on the balanced equation, obtaining 2.00 mol of ClF_3 would require 1.00 mol of Cl_2, and only 0.750 mol of Cl_2 is present. Thus, Cl_2 must be the limiting reactant.
Comment A major point to note from Sample Problems 3.18 and 3.19 is that the relative amounts of reactants do not determine which is limiting, but rather the amount of product formed, which is based on the molar ratio in the balanced equation. In both problems, there is more F_2 than Cl_2. However,
• Sample Problem 3.18 has an F_2/Cl_2 ratio of 6/3, or 2/1, which is less than the required molar ratio of 3/1, so F_2 is limiting and Cl_2 is in excess.
• Sample Problem 3.19 has an F_2/Cl_2 ratio of 3.00/0.750, which is greater than the required molar ratio of 3/1, so Cl_2 is limiting and F_2 is in excess.
Table 1
| Amount (mol) | Cl_2(g) + 3F_2(g) ⟶ 2ClF_3(g) | ||
| Initial | 0.750 | 3.00 | 0 |
| Change | -0.750 | -2.25 | +1.50 |
| Final | 0 | 0.75 | 1.50 |