Calcium carbonate reacts with hydrochloric acid (HCl(aq)) to form calcium chloride and carbon dioxide and water.
a) How much calcium chloride would be formed if 14.7g of calcium carbonate reacted with 250 mL of a 1.6 M solution of hydrochloric acid (HCl(aq)?
b) Calculate the percent yield if 10.6g of calcium chloride was produced.
The Correct Answer and Explanation is :
To solve this problem, we need to follow a systematic approach:
a) How much calcium chloride would be formed if 14.7g of calcium carbonate reacted with 250 mL of a 1.6 M solution of hydrochloric acid (HCl(aq))?
- Write the balanced chemical equation:
[
\text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{CO}_2 (g) + \text{H}_2\text{O} (l)
] - Calculate the moles of calcium carbonate (CaCO₃):
[
\text{Molar mass of CaCO}_3 = 40.08 \, (\text{Ca}) + 12.00 \, (\text{C}) + 3 \times 16.00 \, (\text{O}) = 100.08 \, \text{g/mol}
]
[
\text{Moles of CaCO}_3 = \frac{14.7 \, \text{g}}{100.08 \, \text{g/mol}} = 0.147 \, \text{mol}
] - Calculate the moles of hydrochloric acid (HCl):
[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}
]
[
\text{Moles of HCl} = 1.6 \, \text{M} \times 0.250 \, \text{L} = 0.400 \, \text{mol}
] - Determine the limiting reagent:
From the balanced equation, the mole ratio of CaCO₃ to HCl is 1:2. So, for 0.147 mol of CaCO₃, we need:
[
0.147 \, \text{mol} \times 2 = 0.294 \, \text{mol of HCl}
]
Since we have 0.400 mol of HCl, HCl is in excess, and CaCO₃ is the limiting reagent. - Calculate the moles of calcium chloride (CaCl₂) produced:
From the balanced equation, 1 mol of CaCO₃ produces 1 mol of CaCl₂. So, the moles of CaCl₂ formed are 0.147 mol. - Calculate the mass of calcium chloride (CaCl₂) produced:
The molar mass of CaCl₂ is:
[
\text{Molar mass of CaCl}_2 = 40.08 \, (\text{Ca}) + 2 \times 35.45 \, (\text{Cl}) = 110.98 \, \text{g/mol}
]
[
\text{Mass of CaCl}_2 = 0.147 \, \text{mol} \times 110.98 \, \text{g/mol} = 16.3 \, \text{g}
]
b) Calculate the percent yield if 10.6g of calcium chloride was produced.
- Use the actual yield (10.6g) and the theoretical yield (16.3g) to calculate percent yield:
[
\text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100
]
[
\text{Percent yield} = \frac{10.6 \, \text{g}}{16.3 \, \text{g}} \times 100 = 65.0\%
]
Explanation:
In this reaction, calcium carbonate reacts with hydrochloric acid to produce calcium chloride, carbon dioxide, and water. To determine how much calcium chloride is formed, we first need to identify the limiting reagent. By calculating the moles of each reactant, we find that calcium carbonate is the limiting reagent, so the maximum amount of calcium chloride produced is based on the moles of calcium carbonate. Using stoichiometry, we can convert moles of CaCO₃ to moles of CaCl₂ and then calculate the theoretical mass of CaCl₂.
Next, the percent yield is determined by comparing the actual yield (the amount of CaCl₂ actually produced, 10.6g) to the theoretical yield (the maximum amount that could be produced, 16.3g). A yield of 65.0% indicates that not all of the calcium carbonate reacted efficiently, which could be due to side reactions, incomplete reactions, or losses during handling.