Limiting Reactant and Percent Yield in a Gas-Collection Experiment
A student mixes 5.00 g of calcium carbonate with 50.0 mL of 1.00 M hydrochloric acid in an open flask. The reaction is CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l). The carbon dioxide escapes, and the student collects 0.98 g of CO2 by trapping the gas in a balloon. The student claims calcium carbonate is the limiting reactant because it is the solid and there are 5.00 g of it, while the acid is only 50.0 mL. The instructor asks the class to test that claim with mole calculations, determine the theoretical mass of CO2, and calculate the percent yield. The student also plans a second trial using 100.0 mL of 1.00 M HCl instead of 50.0 mL. The class must predict how the limiting reactant and the theoretical mass of CO2 change in the second trial, and explain why the actual yield is lower than the theoretical yield even when the reaction goes to completion.
1. Show by calculation whether CaCO3 or HCl is limiting. Then calculate the theoretical mass of CO2.
Read the worked answer
Moles of CaCO3: 5.00 g ÷ 100.09 g/mol = 0.04995 mol CaCO3
Moles of HCl: 0.0500 L × 1.00 mol/L = 0.0500 mol HCl
The balanced equation requires 2 mol HCl per 1 mol CaCO3.
Test HCl: 0.0500 mol HCl can react with 0.0500/2 = 0.0250 mol CaCO3. That is less than the 0.04995 mol CaCO3 available, so HCl is limiting.
Theoretical CO2: 0.0500 mol HCl × (1 mol CO2 / 2 mol HCl) = 0.0250 mol CO2.
Mass CO2 = 0.0250 mol × 44.01 g/mol = 1.10 g CO2.
The student's claim is wrong. The solid is not limiting just because it is a solid or because its mass looks large; the mole ratio must be used.
2. Calculate the percent yield using the collected 0.98 g CO2. Give two plausible reasons the yield is below 100% in this open-flask setup, and explain why neither reason changes the limiting reactant.
Read the worked answer
Percent yield: (0.98 g ÷ 1.10 g) × 100% = 89% (about 88.9%).
Two plausible reasons for the loss:
- Some CO2 escaped before the balloon was fully sealed or while the reaction was starting.
- Some CO2 dissolved in the water or was lost during transfer.
A third possibility is that the reaction had not quite gone to completion when collection stopped.
These are experimental losses. They reduce the actual yield but do not change the stoichiometric amount that should form from the initial moles. The limiting reactant is determined by the initial mole amounts and the balanced equation, not by how much product is recovered. Even if all the limiting reactant reacts, the collected product can be less than the theoretical amount.
3. In a second trial, the student uses 100.0 mL of 1.00 M HCl with 5.00 g CaCO3. Identify the limiting reactant and calculate the new theoretical mass of CO2. Explain how the result supports or contradicts the student's original claim that the solid is always limiting.
Read the worked answer
Moles of HCl in trial 2: 0.1000 L × 1.00 mol/L = 0.1000 mol HCl
Moles of CaCO3: still 0.04995 mol.
HCl needed to react with all the CaCO3: 0.04995 mol CaCO3 × 2 = 0.0999 mol HCl
Trial 2 has 0.1000 mol HCl, which is just slightly more than the 0.0999 mol needed. Therefore CaCO3 is limiting in trial 2 (HCl is in very slight excess).
Theoretical CO2: 0.04995 mol CaCO3 × 1 mol CO2/1 mol CaCO3 = 0.04995 mol CO2 0.04995 mol × 44.01 g/mol = 2.20 g CO2
This contradicts the student's original claim that the solid is always limiting. In trial 1, HCl was limiting; in trial 2, CaCO3 is limiting. The limiting reactant depends on the actual mole ratio, which changed when the amount of HCl was doubled.