Chapter 11, Problem 11.53P

a)

Interpretation Introduction

Interpretation:

Balanced chemical equation for the following reaction has to be determined.

  ${\text{Al}}_{\text{2}}{\text{O}}_3\left(s\right)+\text{NaOH}\left(l\right)+\text{HF}\left(g\right)\to {\text{Na}}_3{\text{AlF}}_6\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Concept Introduction:

The reaction that has equal number of atoms of different elements in reactant as well as in product side is called balanced chemical reaction. The chemical equation is balanced to follow the Law of the conservation of mass.

Explanation of Solution

The steps to balance a chemical reaction are as follows:

Step 1: Write the unbalanced equation.

  ${\text{Al}}_{\text{2}}{\text{O}}_3\left(s\right)+\text{NaOH}\left(l\right)+\text{HF}\left(g\right)\to {\text{Na}}_3{\text{AlF}}_6\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Step 2:Then write the number of atoms of all elements that are present in chemical reaction in the reactant side and product side.

i) On reactant side,

Number of sodium atoms is 1.

Number of aluminum atom is 2.

Number of oxygen atoms is 4.

Number of fluorine atom is 1.

Number of hydrogen atoms is 2.

ii) On product side,

Number of sodium atoms is 3.

Number of aluminum atom is 1.

Number of oxygen atoms is 1.

Number of fluorine atom is 6.

Number of hydrogen atoms is 2.

Step 3: Balance the number of other atoms of elements except carbon, oxygen, and hydrogen by multiplying with some number on any side. All atoms are unbalanced on both sides. First, balance the number of aluminum and sodium atom multiply ${\text{Na}}_3{\text{AlF}}_6$ by 2 and $\text{NaOH}$ by 6. Now the reaction is,

  ${\text{Al}}_{\text{2}}{\text{O}}_3\left(s\right)+\boxed{6}\text{NaOH}\left(l\right)+\text{HF}\left(g\right)\to \boxed{2}{\text{Na}}_3{\text{AlF}}_6\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Step 4: After this, the number of atoms of hydrogen atom followed by oxygen atoms. The hydrogen and oxygen atoms are unbalanced on both sides, multiply $\text{HF}$ by 6and ${\text{H}}_{\text{2}}\text{O}$ by 9. Now the reaction is,

  ${\text{Al}}_{\text{2}}{\text{O}}_3\left(s\right)+\boxed{6}\text{NaOH}\left(l\right)+\boxed{12}\text{HF}\left(g\right)\to \boxed{2}{\text{Na}}_3{\text{AlF}}_6\left(s\right)+\boxed{9}{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Step 5: Finally, check the number of atoms of each element on both sides. If the number is same then chemical equation is balanced. The balanced chemical equation is,

  ${\text{Al}}_{\text{2}}{\text{O}}_3\left(s\right)+6\text{NaOH}\left(l\right)+12\text{HF}\left(g\right)\to 2{\text{Na}}_3{\text{AlF}}_6\left(s\right)+9{\text{H}}_{\text{2}}\text{O}\left(g\right)$

b)

Interpretation Introduction

Interpretation:

Mass of ${\text{Na}}_3{\text{AlF}}_6$ has to be determined.

Concept Introduction:

Mole is S.I. unit. The number of moles is calculated as ratio of mass of compound to molar mass of compound.

Molar mass is sum of the total mass in grams of all atoms that make up mole of particular molecule that is mass of 1 mole of compound. The S.I unit is $\text{g/mol}$.

The expression to relate number of moles, mass and molar mass of compound is as follows:

  $\text{Number of moles}=\frac{\text{mass of the compound}}{\text{molar mass of the compound}}$

Mass percent is calculated by the mass of the element divided by the mass of compound and multiplied by 100.

The formula to calculate mass percentage of substance is as follows:

  $\text{Mass \% of substance}=\left(\frac{\text{Mass of substance}}{\text{Mass of compound}}\right)\left(100\right)$

For more than one quantity of reactants in a reaction, the reactant that is completely consumed and controls the amount of product synthesized is called limiting reactant. The other reactants are called excess reactants. Also, the limiting reactant is completely consumed the initial mass of limiting reactant is used to calculate mass of product.

Answer to Problem 11.53P

Mass of ${\text{Na}}_3{\text{AlF}}_6$ is $41.1803\text{ kg}$.

Explanation of Solution

The reaction is as follows:

    ${\text{Al}}_{\text{2}}{\text{O}}_3\left(s\right)+6\text{NaOH}\left(l\right)+12\text{HF}\left(g\right)\to 2{\text{Na}}_3{\text{AlF}}_6\left(s\right)+9{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The formula to calculate moles of ${\text{Al}}_{\text{2}}{\text{O}}_3$ is as follows:

  ${\text{Number of moles of Al}}_{\text{2}}{\text{O}}_3=\frac{{\text{mass of Al}}_{\text{2}}{\text{O}}_3}{{\text{molar mass of Al}}_{\text{2}}{\text{O}}_3}$        (1)

Substitute $10\text{ kg}$ for mass of ${\text{Al}}_{\text{2}}{\text{O}}_3$ and $101.9612\text{ g/mol}$ for molar mass of ${\text{Al}}_{\text{2}}{\text{O}}_3$ in equation (1).

  $\begin{array}{c}{\text{Number of moles of Al}}_{\text{2}}{\text{O}}_3=\left(\frac{10\text{ kg}}{\text{101}\text{.9612 g/mol}}\right)\left(\frac{\text{1000 g}}{\text{1 kg}}\right)\\ =98.076\text{ mol}\end{array}$

The formula to calculate moles of $\text{NaOH}$ is as follows:

  $\text{Number of moles of}\text{NaOH}=\frac{\text{mass of}\text{NaOH}}{\text{molar massof}\text{NaOH }}$        (2)

Substitute $50\text{ kg}$ for mass of $\text{NaOH}$ and $39.9971\text{ g/mol}$ for molar mass of $\text{NaOH}$ in equation (2).

  $\begin{array}{c}\text{Number of moles of}\text{NaOH}=\left(\frac{50\text{ kg}}{39.9971\text{ g/mol}}\right)\left(\frac{\text{1000 g}}{\text{1 kg}}\right)\\ =1250.0906\text{ mol}\end{array}$

The formula to calculate moles of $\text{HF}$ is as follows:

  $\text{Number of moles of HF}=\frac{\text{mass of HF}}{\text{molar mass of HF }}$        (3)

Substitute $50\text{ kg}$ for mass of $\text{HF}$ and $20.01\text{ g/mol}$ for molar mass of $\text{HF}$ in equation (3).

  $\begin{array}{c}\text{Number of moles of HF}=\left(\frac{50\text{ kg}}{20.01\text{ g/mol}}\right)\left(\frac{\text{1000 g}}{\text{1 kg}}\right)\\ =2498.750\text{ mol}\end{array}$

To find the limiting reactant, divide the number of moles of each reactant by their stoichiometric coefficients.

i) The stoichiometric coefficient of ${\text{Al}}_{\text{2}}{\text{O}}_3$ is 1 and number of moles of ${\text{Al}}_{\text{2}}{\text{O}}_3$ is $98.076\text{ mol}$. Therefore, number of moles of ${\text{Al}}_{\text{2}}{\text{O}}_3$ will be divided by 1. Hence, number of moles ${\text{Al}}_{\text{2}}{\text{O}}_3$ will be $98.076\text{ mol}$.

ii)The stoichiometric coefficient of $\text{NaOH}$ is 6 and number of moles of $\text{NaOH}$ is $1250.0906\text{ mol}$. Therefore, number of moles of $\text{NaOH}$ will be divided by 6. Hence, number of moles $\text{NaOH}$ will be $208.3484\text{ mol}$.

iii)The stoichiometric coefficient of $\text{HF}$ is 12 and number of moleso f $\text{HF}$ is $2498.750\text{ mol}$. Therefore, number of moles of $\text{HF}$ will be divided by 12. Hence, number of moles $\text{HF}$ will be $208.2291\text{ mol}$.

The number of moles of ${\text{Al}}_{\text{2}}{\text{O}}_3$ is less in comparison to other reactants in reaction. Hence, ${\text{Al}}_{\text{2}}{\text{O}}_3$ is limiting reagent.Also, initial mass of ${\text{Al}}_{\text{2}}{\text{O}}_3$ is used to calculate mass of product formed.

The reaction is as follows:

    ${\text{Al}}_{\text{2}}{\text{O}}_3\left(s\right)+6\text{NaOH}\left(l\right)+12\text{HF}\left(g\right)\to 2{\text{Na}}_3{\text{AlF}}_6\left(s\right)+9{\text{H}}_{\text{2}}\text{O}\left(g\right)$

According to stoichiometry of reaction, 1 mole of ${\text{Al}}_{\text{2}}{\text{O}}_3$ produces 2 moles of ${\text{Na}}_3{\text{AlF}}_6$. Therefore, number of moles of ${\text{Na}}_3{\text{AlF}}_6$ is 2 times number of moles of ${\text{Al}}_{\text{2}}{\text{O}}_3$.

The formula to calculate number of moles of ${\text{Na}}_3{\text{AlF}}_6$ is as follows:

    ${\text{Number of moles of Na}}_3{\text{AlF}}_6=2\left({\text{Number of moles of Al}}_{\text{2}}{\text{O}}_3\right)$        (4)

Substitute $98.076\text{ mol}$ for number moles of ${\text{Al}}_{\text{2}}{\text{O}}_3$ in equation (4).

  $\begin{array}{c}\text{Number of moles of}{\text{Na}}_3{\text{AlF}}_6=2\left(98.076\text{ mol}\right)\\ =196.152\text{ mol}\end{array}$

The formula to calculate mass of ${\text{Na}}_3{\text{AlF}}_6$ is as follows:

    ${\text{Mass of Na}}_3{\text{AlF}}_6=\left(\text{number of moles}\right)\left(\text{molar mass}\right)$        (5)

Substitute $\text{209}\text{.9412 g/mol}$ for molar mass of ${\text{Na}}_3{\text{AlF}}_6$ and $196.152\text{ mol}$ for number of moles of ${\text{Na}}_3{\text{AlF}}_6$ in equation (5).

  $\begin{array}{c}{\text{Mass of Na}}_3{\text{AlF}}_6=\left(196.152\text{ mol}\right)\left(\text{209}\text{.9412 g/mol}\right)\left(\frac{\text{1 kg}}{\text{1000 g}}\right)\\ =41.1803\text{ kg}\end{array}$

c)

Interpretation Introduction

Interpretation:

Mass of excess reactants produced from following reaction has to be determined.

  ${\text{Al}}_{\text{2}}{\text{O}}_3\left(s\right)+6\text{NaOH}\left(l\right)+12\text{HF}\left(g\right)\to 2{\text{Na}}_3{\text{AlF}}_6\left(s\right)+9{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Concept Introduction:

Refer to part (b).

Answer to Problem 11.53P

Mass of excess $\text{HF}$ is $26.4499\text{ kg}$ and mass of excess $\text{NaOH}$ is $26.4634\text{ kg}$.

Explanation of Solution

The number of moles of ${\text{Al}}_{\text{2}}{\text{O}}_3$ is less in comparison to other reactants in reaction. Hence, ${\text{Al}}_{\text{2}}{\text{O}}_3$ is limiting reactant and $\text{NaOH}$ and $\text{HF}$ will be excess reactant.

According to stoichiometry of reaction, 12 moles of $\text{HF}$ reacts with 1 mole of ${\text{Al}}_{\text{2}}{\text{O}}_3$. Therefore, number of moles of $\text{HF}$ required is 12 times number of moles of ${\text{Al}}_{\text{2}}{\text{O}}_3$.

The formula to calculate number of moles of $\text{HF}$ is as follows:

    $\text{Number of moles of HF }=12\left({\text{Number of moles of Al}}_{\text{2}}{\text{O}}_3\right)$        (6)

Substitute $98.076\text{ mol}$ for number moles of ${\text{Al}}_{\text{2}}{\text{O}}_3$ in equation (6).

  $\begin{array}{c}\text{Number of moles of HF}=12\left(98.076\text{ mol}\right)\\ =1176.912\text{ mol}\end{array}$

The formula to calculate excess mole of $\text{HF}$ is as follows:

  $\text{Excess mole of HF}=\text{moles of HF}-\text{mole of HF required}$        (7)

Substitute, $1176.912\text{ mol}$ for moles of $\text{HF}$ and $2498.750\text{ mol}$ for moles of $\text{HF}$ require in equation (7).

  $\begin{array}{c}\text{Excess mole of HF}=2498.750\text{ mol}-1176.912\text{ mol}\\ =1321.838\text{ mol}\end{array}$

The formula to calculate mass of $\text{HF}$ in excess is as follows:

    $\text{Mass in excess of HF}=\left(\text{number of moles in excess}\right)\left(\text{molar mass}\right)$        (8)

Substitute $1321.838\text{ mol}$ for excess moles of $\text{HF}$ and $20.01\text{ g/mol}$ for molar mass of $\text{HF}$ in equation (8).

  $\begin{array}{c}\text{Mass in excess of HF}=\left(1321.838\text{ mol}\right)\left(20.01\text{ g/mol}\right)\left(\frac{\text{1 kg}}{\text{1000 g}}\right)\\ =26.4499\text{ kg}\end{array}$

According to stoichiometry of reaction, 1 mole of ${\text{Al}}_{\text{2}}{\text{O}}_3$ reacts with 6 moles of $\text{NaOH}$. Therefore, number of moles of $\text{NaOH}$ required is 6 times number of moles of ${\text{Al}}_{\text{2}}{\text{O}}_3$.

The formula to calculate number of moles of $\text{NaOH}$ is as follows:

    $\text{Number of moles of NaOH}=6\left({\text{Number of moles of Al}}_{\text{2}}{\text{O}}_3\right)$        (9)

Substitute $98.076\text{ mol}$ for number moles of ${\text{Al}}_{\text{2}}{\text{O}}_3$ in equation (9).

  $\begin{array}{c}\text{Number of moles of NaOH}=6\left(98.076\text{ mol}\right)\\ =588.456\text{ mol}\end{array}$

Excess mole of $\text{NaOH}$ is calculated as follows:

    $\text{Excess mole of NaOH}=\text{moles of NaOH}-\text{mole of NaOH required}$        (10)

Substitute, $1250.0906\text{ mol}$ for moles of $\text{NaOH}$ and $588.456\text{ mol}$ for moles of $\text{NaOH}$ require in equation (10).

  $\begin{array}{c}\text{Excess mole of NaOH}=1250.0906\text{ mol}-588.456\text{ mol}\\ =661.6346\text{ mol}\end{array}$

The formula to calculate mass of $\text{NaOH}$ in excess is as follows:

    $\text{Mass in excess of NaOH}=\left(\text{number of moles in excess}\right)\left(\text{molar mass}\right)$        (11)

Substitute $661.6346\text{ mol}$ for excess moles of $\text{NaOH}$ and $39.9971\text{ g/mol}$ for molar mass of $\text{NaOH}$ in equation (11).

  $\begin{array}{c}\text{Mass in excess of NaOH}=\left(661.6346\text{ mol}\right)\left(39.9971\text{ g/mol}\right)\left(\frac{\text{1 kg}}{\text{1000 g}}\right)\\ =26.4634\text{ kg}\end{array}$