An excess of 5.0 g of CaCO3 reacts with 100 mL of 0.80 M HCl according to CaCO3 + 2HCl -> CaCl2 + H2O + CO2. What volume of CO2 forms at STP (22.4 L per mole)?
Answer and explanation
Correct answer: A. 0.896 L, set by the limiting HCl
- Step 1: Moles of HCl = 0.100 L x 0.80 M = 0.080 mol; moles of CaCO3 = 5.0 / 100 = 0.050 mol, which needs 0.100 mol HCl, so HCl is limiting.
- Step 2: The 2 : 1 ratio gives 0.080 / 2 = 0.040 mol CO2.
- Step 3: Volume = 0.040 x 22.4 = 0.896 L.
Answer
0.896 L, set by the limiting HCl.
Why not the others: 1.12 L treats CaCO3 as limiting; 1.79 L ignores the 2 : 1 ratio; 0.448 L halves the volume a second time.
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