The number of molecules of Carbon(iv)Oxide produced when 10.0g of CaCO 3 is treated with 0.2dm ^3 of 1 Mole of…

Analyze the question: This is a limiting reagent problem. We must find which reactant runs out first, then use it to find the moles of CO₂ and convert to molecules.

Concept: The reactant that is completely used up (the limiting reagent) decides how much product forms. Number of molecules = moles × Avogadro's number (6.02 × 10²³).

Solution:

1. Moles of CaCO₃: Molar mass of CaCO₃ = 40 + 12 + 48 = 100 g/mol Moles = 10.0 / 100 = 0.10 mol

2. Moles of HCl: "1 mole of HCl" is read as a 1 mol/dm³ solution. Moles = 1 × 0.2 = 0.20 mol

3. Use the equation: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ 1 mol CaCO₃ needs 2 mol HCl. 0.10 mol CaCO₃ needs 0.20 mol HCl. This is exactly what we have, so both react completely (no excess).

4. Moles of CO₂ produced = 0.10 mol (1:1 with CaCO₃)

5. Number of molecules: 0.10 × 6.02 × 10²³ = 6.02 × 10²² molecules

Answer: Option C (6.

Tip: Always compute the moles of each reactant first and compare them using the mole ratio from the balanced equation. Also remember that 0.1 mol of anything contains 6.02 × 10²² particles, so you can often spot the answer quickly.