H for the formation of rust (FeO) is –826 kJ/mol. How much energy is released in the formation of 5 grams of rust?

Question

BlackTom AI · Accepted Answer

First, identify the given data: the enthalpy of formation for rust FeO is ΔH_f = -826 kJ/mol, and we need the energy released when 5.0 g of FeO (rust) forms.

Next, calculate the molar mass of FeO to convert grams to moles. Fe has a molar mass of about 55.85 g/mol and O is about 16.00 g/mol, so M(FeO) ≈ 71.85 g/mol.

Now determine how many moles are in 5.0 g: n = mass / molar mass = 5.0 g / 71.85 g/mol ≈ 0.0696 mol.

Using the enthalpy of formation, the energy change for forming 0.0696 mol is ΔH = n × ΔH_f = 0.0696 mol × (-826 kJ/mol) ≈ -57.5 kJ.

The negative sign indicates energy is released. Thus, the magnitude of energy released for 5.0 g is about 57.5 kJ.

Now evaluate the provided answer choices:
- 25.9 kJ: This is far from the calculated 57.5 kJ; it would correspond to a substantially smaller amount of FeO formed, or a different molar mass or enthalpy value.
- 25.9 J: This is off by several orders of magnitude and incompatible with the 57.5 kJ result; it seems to be a unit or scale error (kJ vs J).
- 66 kJ: This is the closest of the given options to the calculated 57.5 kJ, differing by about 8.5 kJ, which could arise from rounding of molar mass or the enthalpy value.
- 66 J: This is also far from the calculated energy, and inconsistent with the large energy release expected from forming 0.0696 mol of FeO.

In summary, the computed energy release for forming 5.0 g of FeO is approximately 57.5 kJ, so among the choices, the one closest to this value is 66 kJ.

H for the formation of rust (FeO) is –826 kJ/mol. How much energy is released in the formation of 5 grams of rust?单项选择题

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