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 determine the volume of CO₂ gas generated at STP (Standard Temperature and Pressure) from the reaction of 10 g of CaCO₃ with excess HCl, execute the following procedures:

1. Establish 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. Compute 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. Calculate the quantity of moles of CaCO₃ present in 10 g:

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

4. Based on the balanced equation, one mole of CaCO₃ yields one mole of CO₂. Consequently, 0.0999 moles of CaCO₃ will produce 0.0999 moles of CO₂.
5. At STP, a molar volume of 22.4 L is occupied by one mole of any gas. Compute the volume of CO₂ produced:

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

Upon rounding to the correct number of significant figures, the volume is approximately 2.24 L.

Therefore, the volume of CO₂ gas produced at STP is 2.24 L, aligning with the provided selection.