Question

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?

• A. 3.908 g
• B. 2.636 g
• C. 10.545 g
• D. 5.272 g

Hint:

In stoichiometry, make sure the balanced chemical equation is used to relate the moles of reactants and products. Converting moles to grams requires using the molar mass.

Answer 1

The Correct Option is C

The objective is to determine the mass of calcium chloride (\( \text{CaCl}_2 \)) produced from the reaction of 250 mL of 0.76 M hydrochloric acid (\( \text{HCl} \)) with 1000 g of calcium carbonate (\( \text{CaCO}_3 \)). The balanced chemical equation is provided.

Concept Utilized:

This problem involves stoichiometry and identifying the limiting reactant. The limiting reactant is the substance that is entirely consumed during a chemical reaction, thereby dictating the maximum possible yield of the product.

1. Quantify the moles of each reactant based on the given amounts.
2. Employ the stoichiometric ratios from the balanced equation to identify the limiting reactant.
3. Calculate the moles of the product formed, using the moles of the limiting reactant.
4. Convert the moles of the product to its mass using its molar mass.

The formula for calculating moles from molarity and volume is: \( \text{Moles} = \text{Molarity} \times \text{Volume (L)} \). The formula for calculating moles from mass is: \( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \).

Step-by-Step Calculation:

Step 1: Compute the molar masses of the reactants and the product.

Provided 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 for 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: \( \text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l) \).

According to the stoichiometry, 1 mole of \( \text{CaCO}_3 \) reacts with 2 moles of \( \text{HCl} \).

Calculate the moles of \( \text{CaCO}_3 \) required to fully react with the available \( \text{HCl} \):

\[ \text{Moles of } \text{CaCO}_3 \text{ required} = 0.19 \, \text{mol HCl} \times \frac{1 \, \text{mol CaCO}_3}{2 \, \text{mol HCl}} = 0.095 \, \text{mol CaCO}_3 \]

Since 10 moles of \( \text{CaCO}_3 \) are available, which significantly exceeds the 0.095 moles needed, \( \text{CaCO}_3 \) is in excess. Therefore, HCl is the limiting reactant.

Step 4: Calculate the moles of \( \text{CaCl}_2 \) produced.

The quantity of product is determined by the limiting reactant (HCl). The balanced equation indicates that 2 moles of \( \text{HCl} \) yield 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 \( \text{CaCl}_2 \) produced.

Using the molar mass of \( \text{CaCl}_2 \) from 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 is 10.545 g.