Question

When 1.42 g of iron reacts with 1.80 g of chlorine, 3.22 g of $\mathbf{F}\mathbf{e}\mathbf{C}{\mathbf{l}}_{\mathbf{2}}$(s) and 8.60 kJ of heat is produced. What is the enthalpy change for the reaction when 1 mole of $\mathbf{F}\mathbf{e}\mathbf{C}{\mathbf{l}}_{\mathbf{2}}$(s) is produced?

Short answer

Enthalpy change = - 344kJ for 1 mol of $\mathrm{F}\mathrm{e}\mathrm{C}{\mathrm{l}}_2$.

Step-by-step solution

Number of moles

The reaction is:

$2\mathrm{F}\mathrm{e}\left(\mathrm{s}\right)+2\mathrm{C}{\mathrm{l}}_2\left(\mathrm{g}\right)\to \mathrm{F}\mathrm{e}\mathrm{C}{\mathrm{l}}_2\left(\mathrm{s}\right)$

The number of moles is evaluated as:

$\mathbf{N}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{o}\mathbf{f}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}\mathbf{=}\frac{\mathbf{G}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}$

We have

$\frac{1.42\mathrm{g}}{55.8\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}}\times 2=0.04\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{o}\mathrm{f}\mathrm{i}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{i}\mathrm{s}\mathrm{a}\mathrm{v}\mathrm{a}\mathrm{i}\mathrm{l}\mathrm{a}\mathrm{b}\mathrm{l}\mathrm{e}$

$\frac{3.22\mathrm{g}}{126.75\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}}\times 2=0.05\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{o}\mathrm{f}\mathrm{F}\mathrm{e}\mathrm{C}{\mathrm{l}}_2\mathrm{i}\mathrm{s}\mathrm{a}\mathrm{v}\mathrm{a}\mathrm{i}\mathrm{l}\mathrm{a}\mathrm{b}\mathrm{l}\mathrm{e}$

Calculation of enthalpy change

The reaction uses 2 moles of $\mathrm{F}\mathrm{e}\mathrm{C}{\mathrm{l}}_2$and the conversion factor is - 8.60 / 0.05 mol of $\mathrm{F}\mathrm{e}\mathrm{C}{\mathrm{l}}_2$. So we have,

$\begin{array}{l}=\frac{-2\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{F}\mathrm{e}\mathrm{C}{\mathrm{l}}_2\times 8.60}{0.05\mathrm{F}\mathrm{e}\mathrm{C}{\mathrm{l}}_2}\\ =-344\mathrm{k}\mathrm{J}\end{array}$