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A 5.00 g mixture of calcium carbonate (CaCO3, molar mass 100 g/mol)…
Question
A 5.00 g mixture of calcium carbonate (CaCO3, molar mass 100 g/mol) and magnesium carbonate (MgCO3, molar mass 84 g/mol) is heated strongly until decomposition is complete. Each carbonate releases CO2 gas (molar mass 44 g/mol) according to MCO3 -> MO + CO2. The total mass of CO2 evolved is exactly 2.60 g. Calculate the percentage by mass of CaCO3 in the original mixture.
✓ Verified answer: 4.545455checked by our engine — not a guess
Step-by-step solution
Let x = mass of CaCO3 (g), so mass of MgCO3 = (5 - x) g.
Moles CO2 from CaCO3 = x/100; moles CO2 from MgCO3 = (5-x)/84.
Total CO2 mass: 44[x/100 + (5-x)/84] = 2.60.
x/100 + (5-x)/84 = 2.60/44 = 0.0590909...
Multiply by 8400: 84x + 100(5-x) = 0.0590909*8400 = 496.3636
84x + 500 - 100x = 496.3636 -> -16x = -3.6364 -> x = 0.227273 g.
%CaCO3 = 0.227273/5.00 * 100 = 4.5455%.
Final answer4.5454545454545455
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