Chapter 3, Problem 161AP

Potash is any potassium mineral that is used for its potassium content. Most of the potash produced in the United States goes into fertilizer. The major sources of potash are potassium chloride $\left(\text{KCl}\right)$ and potassium sulfate ${\text{(K}}_{\text{2}}{\text{SO}}_{\text{4}}\text{)}$. Potash production is often reported as the potassium oxide ${\text{(K}}_{\text{2}}\text{O)}$ equivalent or the amount of ${\text{K}}_{\text{2}}\text{O}$ that could be made from a given mineral. (a) If $\text{KCl}$ costs $0.55 per kg, for what price (dollar per kg) must ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ be sold to supply the same amount of potassium on a per dollar basis? (b) What mass (in kg) of ${\text{K}}_{\text{2}}\text{O}$ contains the same number of moles of K atoms as 1.00 kg of KG?

Interpretation Introduction

Interpretation:

The price (in dollar/Kg) of potassium sulfate that must be sold to the same supply as of potassiumand the mass (in Kg) of ${\text{K}}_2\text{O}$ required for the given number of potassium atoms is to be calculated.

Concept Introduction:

The atomic mass of elements is measured in terms of the atomic mass unit. One atomic mass unit is defined as the number of elementary particles present in $12.0\text{g}$ of carbon atom.

The number of moles isgiven by the expression, which is as follows:

$n=\frac{\text{given mass}\left(m\right)}{\text{molar mass}\left(M\right)}$.

One mole of a substance contains Avogadro number of particles, that is, $6.022\times {10}^{23}$ particles.

Answer to Problem 161AP

Solution:

a)

$\$0.47\text{ per kg}$

b)

$0.631\text{ kg}$

Explanation of Solution

a) The price (dollar per kg)of ${\text{K}}_2{\text{SO}}_4$.

The cost of $\text{KCl}$ is $\$0.55\text{ per kg}\text{.}$

The molar mass of $\text{KCl}$ is, $M_{\text{KCl}}=74.55\text{ g}/\text{mol}\text{.}$

The molar mass of ${\text{K}}_2{\text{SO}}_4$ is, $M_{{\text{K}}_2{\text{SO}}_4}=174.27\text{ g}/\text{mol}\text{.}$

The molar mass of $\text{K}$ is, $M_{\text{K}}=39.1\text{ g}/\text{mol}\text{.}$

The mass percent can be calculated by the expression, which is as follows:

$\text{mass}\%=\frac{M_{\text{element}}}{M_{\text{compound}}}\times 100\%$

The mass percent of element K in the compound $\text{KCl}$ is calculated as follows:

$\begin{array}{c}\text{mass}\%=\frac{39.1\cancel{\text{g}}/\cancel{\text{mol}}}{74.55\cancel{\text{g}}/\cancel{\text{mol}}}\times 100\%\\ =0.52448\times 100\%\\ =52.45\%.\end{array}$

The mass percent of element K in the compound ${\text{K}}_2{\text{SO}}_4$ is calculated as follows:

$\begin{array}{c}\text{mass}\%=\frac{2\left(39.1\text{ g}/\text{mol}\right)}{174.27\text{ g}/\text{mol}}\times 100\%\\ =\frac{78.2\text{ g}/\text{mol}}{174.27\text{ g}/\text{mol}}\times 100\%\\ =0.4487\times 100\%\\ =44.87\%.\end{array}$

The relation between price and mass percent of potassium is as follows:

$\begin{array}{c}\frac{{\text{Price of K}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{Price of KCl}}=\frac{{\text{\%K in K}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{\%K in KCl}};\\ {\text{Price of K}}_{\text{2}}{\text{SO}}_{\text{4}}=\frac{{\text{\%K in K}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{\%K in KCl}}\times \text{Price of KCl}\text{.}\end{array}$

Substitute $\$0.55\text{ per kg}$ for $\text{Price of KCl}$, $52.45\%$ for ${\text{\%K in K}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $44.87\%$ for $\text{\%K in KCl}$ in the above equation.

$\begin{array}{c}{\text{Price of K}}_{\text{2}}{\text{SO}}_{\text{4}}=\frac{44.87\%}{52.45\%}\times \$0.55\text{ per kg}\\ =0.\text{855}\times \$0.55\text{ per kg}\\ =\$0.47\text{ per kg}\text{.}\end{array}$

b)The mass (in kg) of ${\text{K}}_{\text{2}}\text{O}$ contains the same number of moles of K atoms as 1.00 kg of KCl.

The mass of $\text{KCl}$ is $1.00\text{ kg}$.

The molar mass of $\text{KCl}$ is, $M=74.55\text{ g}/\text{mol}\text{.}$

Convert the mass of $\text{KCl}$ from kilogram to gram, as follows:

$\begin{array}{c}1\text{ kg}=\left(1\cancel{\text{kg}}\right)\left(\frac{1\times {10}^3\text{ g}}{1\cancel{\text{kg}}}\right)\\ =1\times {10}^3\text{ g}\text{.}\end{array}$

The number of moles isgiven by the expression, which is as follows:

$n=\frac{\text{given mass}\left(m\right)}{\text{molar mass}\left(M\right)}$.

Substitute $1\times {10}^3\text{ g}$ for $m$ and $74.55\text{ g}/\text{mol}$ for $M$ to find the number of moles of $\text{KCl}$ in the above equation.

$\begin{array}{c}{n}_{\text{KCl}}=\frac{1\times {10}^3\cancel{\text{g}}}{74.55\cancel{\text{g}}/\text{mol}}\\ =0.0134\times {10}^3\text{ mol}\\ =13.4\text{ mol}\text{.}\end{array}$

In ${\text{K}}_{\text{2}}\text{O,}$ the number of potassium is two whereas in $\text{KCl}$ there is only one potassium. So, the quantity of ${\text{K}}_{\text{2}}\text{O}$ required for $13.4\text{ mol}$ potassium is calculated as follows:

$\begin{array}{c}\text{Mass}\text{of}{\text{K}}_{\text{2}}\text{O}=\frac{13.4\cancel{\text{mol}}}{2}\times 94.2\text{ g}/\cancel{\text{mol}}\times \frac{\text{1 kg}}{{10}^3\text{ g}}\\ \text{=}\frac{\text{1262}\text{.28}\times {\text{10}}^{-3}}{2}\text{ kg}\\ =631.14\times {\text{10}}^{-3}\text{ kg }\\ =0.631\text{ kg}\text{.}\end{array}$.