Consider the following reaction. CaO (s) + CO2 (g) = CaCO3 (s) A chemist allows 14.4 g of CaO and 13.8 g of CO2 to react. When the reaction is finished, the chemist collects 19.4 g of CaCO3. Determine the limiting reactant, theoretical yield, and percent yield for the reactio
options:
CaO; 32 g CaCO3; 60.6%
CaCO3; 25.7 g CaO; 45.5%
CaCO3; 67.2 g CaO; 28.9%
CaO; 25.7 g CaCO3; 75.5%
CaO; 15.0 g CaCO3; 129%
wioeruwqe
To determine the limiting reactant, theoretical yield, and percent yield for the reaction, we need to compare the amounts of reactants and products involved.
Step 1: Calculate the molar masses of CaO, CO2, and CaCO3.
- The molar mass of CaO (calcium oxide) is 40.08 g/mol (1 calcium atom: 40.08 g/mol + 1 oxygen atom: 16 g/mol).
- The molar mass of CO2 (carbon dioxide) is 44.01 g/mol (1 carbon atom: 12.01 g/mol + 2 oxygen atoms: 2 * 16 g/mol).
- The molar mass of CaCO3 (calcium carbonate) is 100.09 g/mol (1 calcium atom: 40.08 g/mol + 1 carbon atom: 12.01 g/mol + 3 oxygen atoms: 3 * 16 g/mol).
Step 2: Convert the given masses of CaO and CO2 into moles.
- Moles of CaO = Mass of CaO / Molar mass of CaO = 14.4 g / 40.08 g/mol = 0.3597 mol.
- Moles of CO2 = Mass of CO2 / Molar mass of CO2 = 13.8 g / 44.01 g/mol = 0.3131 mol.
Step 3: Determine the limiting reactant.
The limiting reactant is the one that is completely consumed in the reaction and determines the amount of product formed. To find the limiting reactant, we compare the mole ratios between the reactants and the product.
The balanced chemical equation for the reaction is:
CaO (s) + CO2 (g) → CaCO3 (s)
From the equation, we see that 1 mole of CaO reacts with 1 mole of CO2 to produce 1 mole of CaCO3.
- The moles ratio between CaO and CaCO3 is 1:1.
- The moles ratio between CO2 and CaCO3 is 1:1.
Since the mole ratio is the same for both reactants, we can compare the moles directly.
Comparing the moles:
- CaO: 0.3597 mol
- CO2: 0.3131 mol
The limiting reactant is CO2 since it is less than CaO.
Step 4: Calculate the theoretical yield.
The theoretical yield is the maximum amount of product that can be obtained from the limiting reactant, assuming complete conversion.
Using the mole ratio from the balanced chemical equation, we can determine the amount of CaCO3 produced from the limiting reactant:
- Moles of CaCO3 from CO2 (limiting reactant) = 0.3131 mol.
To calculate the mass of CaCO3, we multiply the moles by the molar mass of CaCO3:
- Mass of CaCO3 = Moles of CaCO3 from CO2 × Molar mass of CaCO3 = 0.3131 mol × 100.09 g/mol = 31.33 g.
Step 5: Calculate the percent yield.
The percent yield is the ratio of the actual yield (experimental yield) to the theoretical yield, multiplied by 100%.
- Actual yield = 19.4 g (given)
- Percent yield = (Actual yield / Theoretical yield) × 100%
- Percent yield = (19.4 g / 31.33 g) × 100% ≈ 61.89%.
Therefore, the correct option is: CaO; 32 g CaCO3; 60.6%.
afgs
Limiting reagent problem, I think, are easiest to do if you do TWO simple stoichiometry problems. The first time you use CaO and determine moles CaCO3 produced. The second time use 13.8 g CO2 and determine moles CaCO3 produced. The correct answer, in limiting reagent problems, is ALWAYS the smaller value and the reagent producing that smaller value is the limiting reagent. That value of CaCO3 produced will be the theoretical yield.
percent yield = (actual yield/theoretical yield)*100 = ??
(note: actual yield is 19.4 g from the problem.).