Question

Solid iodine trichloride is prepared in two steps: first, a reaction between solid iodine and gaseous chlorine to form iodine monochloride; then, treatment with more chlorine.

(a) Write a balanced equation for each step.

(b) Write a balanced equation for the overall reaction.

(c) How many grams of iodine are needed to prepare 2.45 kg of the final product?

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{The}\mathrm{balanced}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{step}: \\ {\mathbf{I}}_{\mathbf{2}}\left(S\right)\mathbf{+}{\mathbf{Ci}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}\mathbf{ICI}\left(g\right) \\ \mathrm{The}\mathrm{balanced}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{second}\mathrm{step}: \\ \mathbf{ICI}\left(g\right)\mathbf{+}{\mathbf{CI}}_{\mathbf{2}}\left(g\right)\mathbf{\to }{\mathbf{ICI}}_{\mathbf{3}}\left(g\right) \\ \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{}\mathit{T}\mathit{h}\mathit{e}\mathbf{}\mathit{b}\mathit{a}\mathit{l}\mathit{a}\mathit{n}\mathit{c}\mathit{e}\mathit{d}\mathbf{}\mathit{e}\mathit{q}\mathit{u}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathbf{}\mathit{o}\mathit{v}\mathit{e}\mathit{r}\mathit{a}\mathit{l}\mathit{l}\mathbf{}\mathit{r}\mathit{e}\mathit{a}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{}\mathit{i}\mathit{s}\mathbf{:} \\ {\mathbf{I}}_{\mathbf{2}}\mathbf{\left(}\mathbf{S}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{Ci}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{ICI}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ \left(\mathrm{c}\right)\mathrm{The}\mathrm{mass}\mathrm{of}{\mathrm{I}}_2\mathrm{is}1.33\times {10}^3\mathrm{g} \end{gathered}$$

Step-by-step solution

Write a balanced chemical equation of the reaction of solid iodine with gaseous chlorine

When solid iodine $\left(I_2\right)$is treated with gaseous chlorine $\left({\mathrm{CI}}_2\right)$, iodine monochloride $\left(\mathrm{ICI}\right)$is formed. The balanced chemical equation is:

${\mathbf{I}}_{\mathbf{2}}\mathbf{\left(}\mathbf{S}\mathbf{\right)}\mathbf{+}{\mathbf{CI}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{ICI}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{\left(}\mathbf{i}\mathbf{\right)}$

Write a balanced chemical equation of the reaction of iodine monochloride with more chlorine

When iodine monochloride $\left(\mathrm{ICI}\right)$is treated with more chlorine $\left({\mathrm{CI}}_2\right)$iodine trichloride $\left({\mathrm{ICI}}_3\right)$is formed. The balanced chemical equation is:

$\mathbf{ICI}\left(g\right)\mathbf{+}{\mathbf{CI}}_{\mathbf{2}}\left(g\right)\mathbf{\to }{\mathbf{ICI}}_{\mathbf{3}}\left(g\right)\mathbf{\to }\mathbf{\left(}\mathbf{ii}\mathbf{\right)}$

Write a balanced equation of the overall reaction

The balanced overall equation of the reaction is written on the basis of the equation (i) and (ii). For balancing the equation, multiply equation (ii) by 2,

$\mathbf{2}\mathbf{ICI}\left(g\right)\mathbf{+}\mathbf{2}{\mathbf{CI}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}{\mathbf{ICI}}_{\mathbf{3}}\left(g\right)\mathbf{\to }\left(\mathrm{iii}\right)$

The balanced overall equation is defined by adding equations (i) and (iii)

${\mathbf{I}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{CI}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{ICI}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Relation between mass and number of moles

The number of moles is calculated by the mass and Molar mass. The relationship between the number of moles, mass, and molar mass is given below.

$\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{moles}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{M}\mathbf{}\mathbf{olar}\mathbf{}\mathbf{mass}}$

Calculate the number of moles of the final product

$$\begin{gathered} \mathrm{The}\mathrm{mass}\mathrm{of}{\mathbf{ICI}}_{\mathbf{3}}=2.45\mathrm{kg}=2450\mathrm{g}. \\ \mathrm{The}\mathrm{molar}\mathrm{mass}\mathrm{of}\mathbf{}{\mathbf{ICI}}_{\mathbf{3}}=233.2635\mathrm{g}/\mathrm{mol}. \\ \mathrm{The}\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathbf{ICI}}_{\mathbf{3}\mathbf{}}\mathrm{is}\mathrm{calculated}\mathrm{as}: \\ \mathrm{moles}\mathrm{of}{\mathrm{ICI}}_3=\frac{\mathrm{mass}\mathrm{of}\mathrm{I}{\mathrm{CI}}_3}{\mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{ICI}}_3} \\ =\frac{2450\mathrm{g}}{233.2635\mathrm{g}/\mathrm{mol}} \\ \mathrm{Therefore},\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathbf{}{\mathbf{ICI}}_{\mathbf{3}\mathbf{}}\mathrm{is}10.503\mathrm{mol}. \end{gathered}$$

Relation between number of moles of ICI3 and  I2.

In the given reaction, 2mol of ${\mathbf{ICI}}_{\mathbf{3}}$are formed from 1 mol of${\mathbf{I}}_{\mathbf{2}}$.

Thus,

$\mathbf{2}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{ICI}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{I}}_{\mathbf{2}}$

Step 7: Calculate the number of moles of I2

The number of moles of${\mathbf{I}}_{\mathbf{2}}$ is:

$$\begin{gathered} 2\mathrm{mol}\mathrm{of}{\mathrm{ICI}}_3=1\mathrm{mol}\mathrm{of}{\mathrm{I}}_2 \\ 10.503\mathrm{mol}\mathrm{of}{\mathrm{ICI}}_3=10.503\times \frac{1}{2}\mathrm{mol}\mathrm{of}{\mathrm{I}}_2 \\ =5.2515\mathrm{mol}\mathrm{of}{\mathrm{I}}_2 \end{gathered}$$

Calculate the mass of I2

The molar mass of ${\mathbf{I}}_{\mathbf{2}}$= 253.8089 g/mol.

The mass of ${\mathrm{I}}_2$is calculated as:

$$\begin{gathered} \mathrm{mass}\mathrm{of}\mathbf{}{\mathbf{I}}_{\mathbf{2}}=\mathrm{moles}\mathrm{of}\mathbf{}{\mathbf{I}}_{\mathbf{2}}\times \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{I}}_2 \\ =5.2515\mathrm{mol}\times 253.8089\mathrm{g}/\mathrm{mol} \\ =1332.87744\mathrm{g} \end{gathered}$$

Therefore, the mass of ${\mathbf{I}}_{\mathbf{2}}$is approximately$1.33\times {10}^3\mathrm{g}.$