For the reaction of hydrochloric acid and calcium carbonate, the balanced equation is 2 HCl + CaCO3 → CO2 + H2O + CaCl2 If 0.400 mol HCl is reacted with 0.300 mol CaCO3: a. Which reactant is limiting?​​​​​​___ b. How many moles of CO2 can be formed?​​​​___ c. If 0.180 mol of CO2 are actually formed, what is the percent yield? ​ For the reaction of hydrochloric acid and calcium carbonate, the balanced equation is 2 HCl + CaCO3 → CO2 + H2O + CaCl2 If 0.400 mol HCl is reacted with 0.300 mol CaCO3: a. Which reactant is limiting?​​​​​​___ b. How many moles of CO2 can be formed?​​​​___ c. If 0.180 mol of CO2 are actually formed, what is the percent yield? ​____
The Correct Answer and Explanation is :
Reaction:
[ 2 \text{HCl} + \text{CaCO}_3 \rightarrow \text{CO}_2 + \text{H}_2\text{O} + \text{CaCl}_2 ]
Step-by-step solution:
(a) Identify the limiting reactant
- Mole ratio from the balanced equation:
2 mol HCl reacts with 1 mol CaCO₃. - Moles of HCl available: 0.400 mol
- Moles of CaCO₃ available: 0.300 mol
- Calculate how much HCl is needed to react with 0.300 mol CaCO₃:
[
\text{Moles of HCl required} = 0.300 \, \text{mol} \times \frac{2 \, \text{mol HCl}}{1 \, \text{mol CaCO}_3} = 0.600 \, \text{mol HCl}
]
Since only 0.400 mol of HCl is available (less than 0.600 mol), HCl is the limiting reactant.
(b) Moles of CO₂ formed
- Mole ratio from the balanced equation:
2 mol HCl produces 1 mol CO₂. - Calculate moles of CO₂ produced based on the limiting reactant (HCl):
[
\text{Moles of CO}_2 = 0.400 \, \text{mol HCl} \times \frac{1 \, \text{mol CO}_2}{2 \, \text{mol HCl}} = 0.200 \, \text{mol CO}_2
]
(c) Percent yield
- Actual moles of CO₂ formed: 0.180 mol
- Theoretical moles of CO₂ formed: 0.200 mol
- Percent yield calculation:
[
\text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 = \left( \frac{0.180}{0.200} \right) \times 100 = 90\%
]
Final Answers
a. Limiting reactant: HCl
b. Moles of CO₂ formed: 0.200 mol
c. Percent yield: 90%
Explanation (300 words):
In this reaction, hydrochloric acid (HCl) reacts with calcium carbonate (CaCO₃) to produce carbon dioxide (CO₂), water (H₂O), and calcium chloride (CaCl₂). To solve the problem, the first step is to identify the limiting reactant. The balanced chemical equation shows a 2:1 molar ratio between HCl and CaCO₃. With 0.400 mol of HCl and 0.300 mol of CaCO₃ available, we calculate that 0.300 mol of CaCO₃ would require 0.600 mol of HCl. However, since only 0.400 mol of HCl is present, it limits the reaction.
Using the limiting reactant (HCl), we determine how much CO₂ is produced. According to the stoichiometry, 2 mol of HCl produces 1 mol of CO₂. Therefore, 0.400 mol of HCl generates 0.200 mol of CO₂ under ideal conditions (theoretical yield).
Finally, the percent yield is calculated to evaluate the reaction efficiency. The actual amount of CO₂ formed is 0.180 mol, and dividing this by the theoretical yield (0.200 mol), then multiplying by 100, gives a percent yield of 90%. This indicates that 90% of the expected product was obtained, reflecting some loss or inefficiency in the reaction.