Question

iron(iii) oxide is formed when pure iron reacts with oxygen according to the following thermochemical equation:

4 fe(s) + 3 o₂(g) → 2 fe₂o₃(s); δh = -1648.4 kj

how much heat is released or absorbed (given by the sign of your answer) when 6.001 g of iron are reacted in excess oxygen? report your answer in kilojoules to 4 sig figs, but do not include the units in the answer.

Explanation:

Step1: Calculate the number of moles of Fe

The molar mass of Fe (\(M_{Fe}\)) is \(55.845\ g/mol\).
The number of moles of Fe (\(n_{Fe}\)) is calculated using the formula \(n=\frac{m}{M}\), where \(m = 6.001\ g\) and \(M = 55.845\ g/mol\).
\(n_{Fe}=\frac{6.001}{55.845}\ mol\)

Step2: Relate moles of Fe to \(\Delta H\)

From the thermochemical equation \(4Fe_{(s)}+3O_{2(g)}\to2Fe_{2}O_{3(s)};\Delta H=- 1648.4\ kJ\), when \(4\ mol\) of Fe reacts, the heat change is \(\Delta H=-1648.4\ kJ\).
Let \(x\) be the heat change when \(n_{Fe}=\frac{6.001}{55.845}\ mol\) of Fe reacts.
We set up a proportion: \(\frac{x}{\frac{6.001}{55.845}}=\frac{-1648.4}{4}\)

$$x=\frac{-1648.4\times6.001}{4\times55.845}$$

Step3: Calculate the value of \(x\)

$$x=\frac{-1648.4\times6.001}{4\times55.845}=\frac{-9891.0484}{223.38}\approx - 44.28$$

Answer:

\(-44.28\)