Calcium carbonate reacts with aqueous HCl to give CaCl2and CO2according to the reaction given below: CaCO3(s) + 2HCl (aq) ? CaCl2(aq) + CO2(g) + H2O(l) What mass of CaCl2will be formed when 250 mL of 0.76 M HCI reacts with 1000 g of CaCO3? Name the limiting reagent. Calculate the number of moles of CaCl2formed in the reaction.
The Correct Answer and Explanation is :
To determine the mass of CaCl₂ formed in this reaction, we first need to establish the limiting reagent and then calculate the amount of CaCl₂ produced.
Step 1: Determine the moles of HCl and CaCO₃
The balanced chemical equation is:
[
\text{CaCO₃ (s)} + 2\text{HCl (aq)} \rightarrow \text{CaCl₂ (aq)} + \text{CO₂ (g)} + \text{H₂O (l)}
]
Moles of HCl:
The concentration of HCl is 0.76 M (moles per liter), and the volume of HCl is 250 mL (0.250 L). The number of moles of HCl is calculated using the formula:
[
\text{moles of HCl} = \text{Molarity (M)} \times \text{Volume (L)}
]
[
\text{moles of HCl} = 0.76 \, \text{M} \times 0.250 \, \text{L} = 0.19 \, \text{moles of HCl}
]
Moles of CaCO₃:
The molar mass of CaCO₃ is approximately 100.1 g/mol. The number of moles of CaCO₃ is:
[
\text{moles of CaCO₃} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{1000 \, \text{g}}{100.1 \, \text{g/mol}} \approx 9.99 \, \text{moles of CaCO₃}
]
Step 2: Determine the limiting reagent
From the balanced equation, 1 mole of CaCO₃ reacts with 2 moles of HCl. Therefore, to fully react with 9.99 moles of CaCO₃, we would need:
[
9.99 \, \text{moles of CaCO₃} \times 2 \, \text{moles of HCl} = 19.98 \, \text{moles of HCl}
]
However, we only have 0.19 moles of HCl, which is far less than the 19.98 moles required. Therefore, HCl is the limiting reagent.
Step 3: Calculate the moles of CaCl₂ produced
From the balanced equation, 2 moles of HCl produce 1 mole of CaCl₂. Since HCl is the limiting reagent, we can calculate the moles of CaCl₂ formed based on the available moles of HCl:
[
\text{moles of CaCl₂} = \frac{0.19 \, \text{moles of HCl}}{2} = 0.095 \, \text{moles of CaCl₂}
]
Step 4: Calculate the mass of CaCl₂
The molar mass of CaCl₂ is approximately 147 g/mol. The mass of CaCl₂ produced is:
[
\text{mass of CaCl₂} = \text{moles of CaCl₂} \times \text{molar mass of CaCl₂}
]
[
\text{mass of CaCl₂} = 0.095 \, \text{moles} \times 147 \, \text{g/mol} = 13.9 \, \text{g of CaCl₂}
]
Conclusion
- The limiting reagent is HCl.
- The number of moles of CaCl₂ formed is 0.095 moles.
- The mass of CaCl₂ formed is 13.9 g.
This shows that HCl limits the amount of CaCl₂ produced, even though there is excess CaCO₃.