Chapter 6, Problem 6.90QE

Interpretation Introduction

Interpretation:

The mass of sodium used in a reaction when $499\text{mL}$ of wet hydrogen is collected over water has to be calculated.

Concept Introduction:

In any stoichiometric balanced equation, the reactant that is present in the least amount and that governs the amount of products formed is known as limiting reactant.

Prior to the identification of the limiting reactant; a balanced equation is written, then the amount of each reactant in grams is converted to corresponding moles. Further on the basis of the stoichiometric molar ratio by which reactants combine, the amount of product formed from each of the reactants is calculated.

Answer to Problem 6.90QE

The mass of sodium used in a reaction is $0.91402\text{ g}$.

Explanation of Solution

The conversion factor to convert liters to milliliters is as follows:

  $\text{1 L}=\text{1000 mL}$

Hence convert $499\text{ mL}$ to $\text{L}$ as follows:

  $\begin{array}{c}{\text{Volume of H}}_2=\left(499\text{}\text{mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =0.499\text{ L}\end{array}$

The barometric pressure is the total pressure measured and the vapor pressure of the water at $22°\text{C}$ is $22\text{ torr}$.

The total pressure as per Dalton’s law of partial pressure is calculated as follows:

  $P_{\text{T}}=P_{{\text{H}}_{\text{2}}\text{O}}+P_{{\text{H}}_{\text{2}}}$        (1)

Here,

$P_{\text{T}}$ denotes the total pressure exerted by the mixture of gases.

$P_{{\text{H}}_{\text{2}}\text{O}}$ denotes the partial pressure exerted by the water vapor.

$P_{{\text{H}}_{\text{2}}}$ denotes the partial pressure exerted by the ${\text{H}}_{\text{2}}$.

Substitute $22\text{ torr}$ for $P_{{\text{H}}_{\text{2}}\text{O}}$, $755\text{ torr}$ for $P_{\text{T}}$ in equation (1).

  $755\text{ torr}=22\text{ torr}+P_{{\text{H}}_{\text{2}}}$

Rearrange the expression to obtain the value of $P_{{\text{H}}_{\text{2}}}$.

  $\begin{array}{c}{P}_{{\text{H}}_{\text{2}}}=755\text{ torr}-\text{22 torr}\\ =\text{733 torr}\end{array}$

Therefore the partial pressure corresponding to ${\text{H}}_{\text{2}}$ is $\text{733 torr}$.

The conversion factor to convert $\text{atm}$ to $\text{torr}$ is as follows:

  $1\text{ atm}=\text{760 torr}$

Therefore $\text{733 torr}$ is converted to $\text{atm}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(\text{733 torr}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\\ =0.964\text{ atm}\end{array}$

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$        (2)

Substitute $\text{22 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (2).

  $\begin{array}{c}T\left(\text{K}\right)=22°\text{C}+273\text{ K}\\ =295\text{ K}\end{array}$

The formula to calculate the number of moles as per the ideal gas equation is as follows:

  $n=\frac{PV}{RT}$        (3)

Substitute $0.499\text{ L}$ for $V$, $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for gas constant $R$, $\text{295 K}$ for temperature and $\text{0}\text{.964 atm}$ for $P$ in equation (3).

  $\begin{array}{c}n=\frac{\left(\text{0}\text{.964 atm}\right)\left(0.499\text{ L}\right)}{\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(\text{295 K}\right)}\\ =1.987\times {10}^{-2}\text{mol}\end{array}$

The balanced reaction when sodium reacts with water is given as follows:

  $2\text{Na}+2{\text{H}}_{\text{2}}\text{O}\to 2\text{NaOH}+{\text{H}}_{\text{2}}$        (4)

According to the stoichiometry of the balanced equation (4), $\text{2 mol Na}$ is required to produce one mole of ${\text{H}}_{\text{2}}$, therefore the number of moles of $\text{Na}$ required to produce $1.987\times {10}^{-2}{\text{mol H}}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}\text{Number of moles of Na}=\left(1.987\times {10}^{-2}{\text{mol H}}_{\text{2}}\right)\left(\frac{{\text{2 mol Na}}_{\text{2}}}{{\text{1 mol H}}_{\text{2}}}\right)\\ =3.974\times {10}^{-2}\text{mol Na}\end{array}$

The formula to convert mass in gram to moles is as follows:

  $\text{Mass}=\text{Number of moles}\times \text{molar mass}$        (5)

Substitute $3.974\times {10}^{-2}\text{mol}$ for the number of moles and $\text{23 g/mol}$ for molar mass to calculate the mass of sodium in equation (5).

  $\begin{array}{c}\text{Mass}=\left(3.974\times {10}^{-2}\text{mol}\right)\left(\text{23 g/mol}\right)\\ =0.91402\text{ g}\end{array}$