Chapter 12, Problem 12.71E

Interpretation Introduction

Interpretation:

The limiting reactant in the reaction of $\text{25}\text{.0}\text{g}$ of $\text{2-butene}$ with $\text{25}\text{.0}\text{g}$ of iodine is to be predicted. The mole of product produced in the reaction of $\text{25}\text{.0}\text{g}$ of $\text{2-butene}$ reacts with $\text{25}\text{.0}\text{g}$ of iodine is to be predicted.

Concept Introduction:

The reactant that determines the concentration of the product formed in a chemical reaction is known as limiting reagent. The other reactants left in the reaction mixture are termed as reactants in excess.

Answer to Problem 12.71E

The limiting reactant in the reaction of $\text{25}\text{.0}\text{g}$ of $\text{2-butene}$ with $\text{25}\text{.0}\text{g}$ of iodine is ${\text{I}}_2$. The number of moles of $\text{2, 3-diiodobutane}$ are $0.0985\text{mol}$.

Explanation of Solution

The reaction of butane with iodine results in the formation of $\text{2, 3-diiodobutane}$. The corresponding chemical reaction is shown below.

Figure 1

The molecular formula of $\text{2-butene}$ is ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}$. The molar mass of carbon and hydrogen is $\text{12}\text{g}/\text{mol}$ and $\text{1}\text{g}/\text{mol}$ respectively.

The molar mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}$ is calculated as follow:

$\begin{array}{rll}{\text{Molar mass of C}}_{\text{4}}{\text{H}}_{\text{8}}& =& \left(4\times \text{mass of C atom + 8}\times \text{mass of H atom}\right)\\ & =& \left(4\times 12.01+8\times 1.008\right)\\ & =& \left(48.04+8.064\right)\\ & =& \text{56}\text{.104 g/mol}\end{array}$

The molar mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}$ is $\text{56}\text{.104 g/mol}$. The mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}$ is given as $\text{25}\text{.0}\text{g}$.

The number of moles is calculated by the formula shown below.

$n=\frac{\text{Given}\text{Mass}}{\text{Molar}\text{Mass}}$ …(1)

Substitute molar mass of and given mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}$ in equation (1).

$\begin{array}{rll}n& =& \frac{\text{25}\text{.0 g}}{\text{56}\text{.104 g/mol}}\\ & =& 0.4456\text{mol}\end{array}$

Therefore, the number of moles of ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}$ is $0.4456\text{mol}$.

The molar mass of iodine is $\text{126}\text{.90}\text{g}/\text{mol}$.

The molar mass of ${\text{I}}_2$ is calculated as follow:

$\begin{array}{rll}{\text{Molar mass of I}}_2& =& 2\times \text{mass of I atom}\\ & =& 2\times \text{126}\text{.90}\text{g}/\text{mol}\\ & =& \text{253}\text{.8 g/mol}\end{array}$

The molar mass of ${\text{I}}_2$ is $\text{253}\text{.8 g/mol}$. The mass of ${\text{I}}_2$ is given as $\text{25}\text{.0}\text{g}$.

Substitute molar mass of and given mass of ${\text{I}}_2$ in equation (1).

$\begin{array}{rll}n& =& \frac{\text{25}\text{.0 g}}{\text{253}\text{.8 g/mol}}\\ & =& 0.0985\text{mol}\end{array}$

Therefore, the number of moles of ${\text{I}}_2$ is $0.0985\text{mol}$.

The number of moles of ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}$ is greater than the number of moles of ${\text{I}}_2$. Therefore, the limiting reactant is ${\text{I}}_2$. Therefore, the $0.0985\text{mol}$ of ${\text{I}}_2$ produces $0.0985\text{mol}$ of $\text{2, 3-diiodobutane}$.

Conclusion

The limiting reactant in the reaction of $\text{25}\text{.0}\text{g}$ of $\text{2-butene}$ with $\text{25}\text{.0}\text{g}$ of iodine is ${\text{I}}_2$. The number of moles of $\text{2, 3-diiodobutane}$ are $0.0985\text{mol}$.