Question

If \(20 \text{ g}\) of \(\text{CaCO}_3\) is treated with \(100 \text{ mL}\) of \(20\%\) \(\text{HCl}\) solution, the amount of \(\text{CO}_2\) produced is:

• A. \(22.4 \text{ L}\)
• B. \(8.80 \text{ g}\)
• C. \(4.40 \text{ g}\)
• D. \(2.24 \text{ L}\)

Hint:

Always double check the units given in the options! Options (2) and (3) are given in grams (g), while options (1) and (4) are in liters (L). Calculating the mass first saves you from doing an unnecessary conversion to volume.

Answer 1

The Correct Option is B

Concept: This problem requires a stoichiometric calculation involving a limiting reagent. First, we write the balanced chemical equation representing the reaction between calcium carbonate and hydrochloric acid: \[ \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \] From the balanced equation, \(1 \text{ mole}\) of \(\text{CaCO}_3\) reacts completely with \(2 \text{ moles}\) of \(\text{HCl}\) to yield \(1 \text{ mole}\) of \(\text{CO}_2\). We must find the initial moles of each reactant to determine which one limits the production of products. Step 1: Calculate the moles of \(\text{CaCO}_3\) available.
The molar mass of \(\text{CaCO}_3\) is: \[ \text{Molar Mass} = 40 (\text{Ca}) + 12 (\text{C}) + 3 \times 16 (\text{O}) = 100 \text{ g/mol} \] Given mass of \(\text{CaCO}_3 = 20 \text{ g}\): \[ \text{Moles of }\text{CaCO}_3 = \frac{20 \text{ g}}{100 \text{ g/mol}} = 0.2 \text{ moles} \]

Step 2: Calculate the moles of \(\text{HCl}\) available.
Assuming the \(20\%\) \(\text{HCl}\) solution is specified by mass/volume percentage (% w/v), which is typical for standard aqueous reactions in introductory chemistry unless otherwise specified: \[ 20\% \text{ w/v means } 20 \text{ g of HCl is present in } 100 \text{ mL of solution.} \] Therefore, the mass of \(\text{HCl}\) in our \(100 \text{ mL}\) solution is exactly \(20 \text{ g}\).
The molar mass of \(\text{HCl}\) is: \[ \text{Molar Mass} = 1 (\text{H}) + 35.5 (\text{Cl}) = 36.5 \text{ g/mol} \] Calculating the moles of \(\text{HCl}\): \[ \text{Moles of }\text{HCl} = \frac{20 \text{ g}}{36.5 \text{ g/mol}} \approx 0.548 \text{ moles} \]

Step 3: Determine the limiting reagent.
According to stoichiometry, \(0.2 \text{ moles}\) of \(\text{CaCO}_3\) requires: \[ 2 \times 0.2 = 0.4 \text{ moles of HCl} \] Since we possess \(0.548 \text{ moles}\) of \(\text{HCl}\) (which is greater than \(0.4 \text{ moles}\)), \(\text{HCl}\) is present in excess. Thus, \(\text{CaCO}_3\) is the limiting reagent and dictates the maximum amount of \(\text{CO}_2\) formed.

Step 4: Calculate the mass of \(\text{CO}_2\) produced.
The stoichiometric ratio shows that \(1 \text{ mole}\) of \(\text{CaCO}_3\) generates \(1 \text{ mole}\) of \(\text{CO}_2\). Therefore, \(0.2 \text{ moles}\) of \(\text{CaCO}_3\) yields \(0.2 \text{ moles}\) of \(\text{CO}_2\).
The molar mass of \(\text{CO}_2\) is: \[ \text{Molar Mass} = 12 (\text{C}) + 2 \times 16 (\text{O}) = 44 \text{ g/mol} \] Calculating the final mass: \[ \text{Mass of }\text{CO}_2 = 0.2 \text{ moles} \times 44 \text{ g/mol} = 8.80 \text{ g} \]