How many moles of carbon are present in 125.0 g of iron(III) carbonate (Fe2(CO3)3)?

• A. 1.14 mol
• B. 0.72 mol
• C. 2.28 mol
• D. 0.36 mol

Answer: (A)

Verified step by step guidance

1. Identify the chemical formula of iron(III) carbonate, which is \(\mathrm{Fe_2(CO_3)_3}\), and note that each formula unit contains 3 carbonate ions, each with 1 carbon atom, so there are 3 carbon atoms per formula unit.
2. Calculate the molar mass of iron(III) carbonate by summing the atomic masses of all atoms in the formula: 2 iron (Fe) atoms, 3 carbon (C) atoms, and 9 oxygen (O) atoms. Use atomic masses approximately Fe = 55.85 g/mol, C = 12.01 g/mol, and O = 16.00 g/mol.
3. Determine the number of moles of iron(III) carbonate in 125.0 g by dividing the given mass by the molar mass calculated in the previous step: \(\text{moles of } \mathrm{Fe_2(CO_3)_3} = \frac{125.0\ \mathrm{g}}{\text{molar mass of } \mathrm{Fe_2(CO_3)_3}}\).
4. Calculate the moles of carbon atoms by multiplying the moles of iron(III) carbonate by the number of carbon atoms per formula unit: \(\text{moles of C} = \text{moles of } \mathrm{Fe_2(CO_3)_3} \times 3\).
5. The result from the previous step gives the total moles of carbon present in 125.0 g of iron(III) carbonate.