Consider the following reaction:
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) ΔH = –1652 kJ

How much heat is released when 31.0 g Fe(s) is reacted with excess O2(g)?
(MM of Fe = 55.85 g/mol)

Do NOT include the positive or negative sign in your answer!

Round your answer to ZERO decimal places (i.e. 274) and do NOT include the units of kJ.

03A

Question

Consider the following reaction:
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)     ΔH = –1652 kJ

How much heat is released when 31.0 g Fe(s) is reacted with excess O2(g)?
(MM of Fe = 55.85 g/mol)

Do NOT include the positive or negative sign in your answer!

Round your answer to ZERO decimal places (i.e. 274) and do NOT include the units of kJ.

03A

BlackTom AI · Accepted Answer

First, restate the given data and what is being asked: the reaction releases heat with ΔH = –1652 kJ for the combustion of 4 moles of Fe. We need to find the heat released when 31.0 g of Fe reacts with excess O2, and we will report the magnitude without a sign and without units, rounded to zero decimals.

Step 1: Convert the mass of iron to moles. The molar mass of Fe is 55.85 g/mol, so moles of Fe = 31.0 g ÷ 55.85 g/mol ≈ 0.5549 mol.

Step 2: Use the stoichiometry of the reaction. The enthalpy change corresponds to 4 moles of Fe reacting. Therefore, the heat per mole of Fe is ΔH per 4 mol Fe = (–1652 kJ) ÷ 4 = –413 kJ/mol Fe.

Step 3: Multiply the moles of Fe by the heat per mole to get the total heat: (0.5549 mol) × (–413 kJ/mol) ≈ –229 kJ.

Step 4: Present the required form. The problem asks for the heat released without a sign and without units, rounded to zero decimal places. The magnitude is about 229 kJ, so the numeric value to report is 229.

Note on rounding and sign convention: since the question requests no sign, we take the absolute value of the heat and round as specified. If you were to include units, the heat would be approximately 229 kJ (released). The essential checks are ensuring proper molar conversion, correct stoichiometric scaling from 4 Fe to the given amount, and careful rounding to zero decimals.

CHEM 1210 AU2025 (15738) Exam 3 (Make-up)- Requires Respondus LockDown Browser

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