Question

What volume of CO$_2$ gas at STP is produced by the reaction of 10 g of CaCO$_3$ with excess HCl?

• A. 4.48 L
• B. 2.24 L
• C. 8.96 L
• D. 11.2 L

Hint:

At STP, use the relation: Volume = Moles × 22.4 L. Always verify molar mass and units.

Answer 1

The Correct Option is B

To find the volume of CO₂ gas produced at STP (Standard Temperature and Pressure) when 10 g of CaCO₃ reacts with excess HCl, follow these steps:

1. Start with the balanced chemical equation for the reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl):

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

2. Calculate the molar mass of CaCO₃:

Molar mass of CaCO₃ = (40.08 g/mol for Ca) + (12.01 g/mol for C) + (3 × 16.00 g/mol for O)
Molar mass of CaCO₃ = 100.09 g/mol

3. Determine the number of moles of CaCO₃ in 10 g:

Number of moles = 10 g / 100.09 g/mol = 0.0999 mol

4. According to the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, 0.0999 moles of CaCO₃ will produce 0.0999 moles of CO₂.
5. At STP, 1 mole of gas occupies 22.4 L. Calculate the volume of CO₂ produced:

Volume of CO₂ = 0.0999 mol × 22.4 L/mol = 2.2368 L

Rounding to appropriate significant figures, the volume is approximately 2.24 L.

Therefore, the volume of CO₂ gas produced at STP is 2.24 L, which matches the given option.

Answer 2

Step 1: Write the balanced equation
\[ \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \] Step 2: Calculate moles of CaCO$_3$
\[ \text{Moles of CaCO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{10\ \text{g}}{100\ \text{g/mol}} = 0.1\ \text{mol} \] Step 3: Determine moles of CO$_2$ produced
From the 1:1 mole ratio:
\[ \text{Moles of CO}_2 = 0.1\ \text{mol} \] Step 4: Calculate volume at STP
\[ \text{Volume} = \text{Moles} \times \text{Molar volume} = 0.1\ \text{mol} \times 22.4\ \text{L/mol} = 2.24\ \text{L} \] The correct volume of CO$_2$ produced is $\boxed{2.24\ \text{L}}$.