Consider the above reaction, what mass of CaCl₂ will be formed if 250 ml of 0.76 M HCl reacts with 1000 g of CaCO₃?
Consider the above reaction, what mass of CaCl₂ will be formed if 250 ml of 0.76 M HCl reacts with 1000 g of CaCO₃?
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- 3.908 g
- 2.636 g
- 10.545 g
- 5.272 g
The Correct Option is C
Approach Solution - 1
We are asked to calculate the mass of calcium chloride (\( \text{CaCl}_2 \)) formed from the reaction between 250 mL of 0.76 M hydrochloric acid (\( \text{HCl} \)) and 1000 g of calcium carbonate (\( \text{CaCO}_3 \)). The balanced chemical equation is provided.
Concept Used:
This is a stoichiometry problem that involves a limiting reactant. The limiting reactant is the reactant that is completely consumed in a chemical reaction and therefore determines the maximum amount of product that can be formed.
- Calculate the moles of each reactant from the given quantities.
- Use the stoichiometry of the balanced equation to determine which reactant is the limiting one.
- Calculate the moles of the product formed based on the moles of the limiting reactant.
- Convert the moles of the product to mass using its molar mass.
The formula for moles from molarity and volume is: \( \text{Moles} = \text{Molarity} \times \text{Volume in Liters} \). The formula for moles from mass is: \( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \).
Step-by-Step Solution:
Step 1: Calculate the molar masses of the reactants and the product.
Given molar masses: Ca = 40, C = 12, O = 16, Cl = 35.5 g/mol.
- Molar mass of \( \text{CaCO}_3 = 40 + 12 + 3(16) = 100 \, \text{g/mol} \).
- Molar mass of \( \text{CaCl}_2 = 40 + 2(35.5) = 40 + 71 = 111 \, \text{g/mol} \).
Step 2: Calculate the number of moles of each reactant.
For HCl:
\[ \text{Volume} = 250 \, \text{mL} = 0.250 \, \text{L} \] \[ \text{Molarity} = 0.76 \, \text{M} = 0.76 \, \text{mol/L} \] \[ \text{Moles of HCl} = 0.76 \, \frac{\text{mol}}{\text{L}} \times 0.250 \, \text{L} = 0.19 \, \text{mol} \]
For \( \text{CaCO}_3 \):
\[ \text{Mass} = 1000 \, \text{g} \] \[ \text{Molar Mass} = 100 \, \text{g/mol} \] \[ \text{Moles of } \text{CaCO}_3 = \frac{1000 \, \text{g}}{100 \, \text{g/mol}} = 10 \, \text{mol} \]
Step 3: Identify the limiting reactant.
The balanced chemical equation is: \( \text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l) \).
The stoichiometry shows that 1 mole of \( \text{CaCO}_3 \) reacts with 2 moles of \( \text{HCl} \).
Let's determine how many moles of \( \text{CaCO}_3 \) are required to react completely with the available \( \text{HCl} \):
\[ \text{Moles of } \text{CaCO}_3 \text{ needed} = 0.19 \, \text{mol HCl} \times \frac{1 \, \text{mol CaCO}_3}{2 \, \text{mol HCl}} = 0.095 \, \text{mol CaCO}_3 \]
We have 10 moles of \( \text{CaCO}_3 \) available, which is much more than the 0.095 moles needed. Therefore, \( \text{CaCO}_3 \) is in excess, and HCl is the limiting reactant.
Step 4: Calculate the moles of \( \text{CaCl}_2 \) formed.
The amount of product formed is determined by the limiting reactant (HCl). From the balanced equation, 2 moles of \( \text{HCl} \) produce 1 mole of \( \text{CaCl}_2 \).
\[ \text{Moles of } \text{CaCl}_2 = 0.19 \, \text{mol HCl} \times \frac{1 \, \text{mol CaCl}_2}{2 \, \text{mol HCl}} = 0.095 \, \text{mol CaCl}_2 \]
Step 5: Calculate the mass of \( \text{CaCl}_2 \) formed.
Using the molar mass of \( \text{CaCl}_2 \) calculated in Step 1:
\[ \text{Mass of } \text{CaCl}_2 = \text{Moles} \times \text{Molar Mass} \] \[ \text{Mass of } \text{CaCl}_2 = 0.095 \, \text{mol} \times 111 \, \text{g/mol} = 10.545 \, \text{g} \]
The mass of \( \text{CaCl}_2 \) formed will be 10.545 g.
Approach Solution -2
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