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

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.

1. Calculate the moles of each reactant from the given quantities.
2. Use the stoichiometry of the balanced equation to determine which reactant is the limiting one.
3. Calculate the moles of the product formed based on the moles of the limiting reactant.
4. 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.

Answer 2

Using stoichiometry, the moles of \( \text{CaCO}_3 \) are calculated: \[ \text{Moles of CaCO}_3 = \frac{1000}{100} = 10 \, \text{mol} \] Now, for the reaction: \[ \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \] Hence, the moles of \( \text{CaCl}_2 \) formed will be \( 10 \, \text{mol} \). Finally: \[ \text{Mass of CaCl}_2 = 10.545 \, \text{g} \]