5.00 g of calcium carbonate (M = 100.1 g/mol) reacts with excess hydrochloric acid: CaCO3 + 2HCl -> CaCl2 + H2O + CO2. What volume of CO2 forms at SLC (molar volume 24.8 L/mol)?
Answer and explanation
Correct answer: D. 1.24 L
- Step 1: n(CaCO3) = 5.00 / 100.1 = 0.0500 mol.
- Step 2: The ratio CaCO3 : CO2 is 1 : 1, so n(CO2) = 0.0500 mol.
- Step 3: V = 0.0500 x 24.8 = 1.24 L.
Answer
1.24 L.
Why not the others: 2.48 L doubles it using the HCl ratio; 0.620 L halves it; 4.95 L divides by the molar mass wrongly.
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