Chapter 6, Problem 6.64QE

Interpretation Introduction

Interpretation:

Volume of hydrogen gas produced when $3.43\text{ g}$ of $\text{Fe}$ metal reacts with $40.0\text{ mL}$ of $2.43M\text{ HCl}$ has to be calculated.

Concept Introduction:

Any gas obeys the assumption laid down in kinetic molecular theory is said to be an ideal gas. The combination of all the gas laws namely Boyle’s law, Charles law and Avogadro’s leads to the ideal gas equation that can be used to relate all the four properties such as given below:

  $PV=nRT$

Here,

$\text{R}$ is the universal gas constant.

$V$ denotes the volume.

$n$ denotes the number of moles.

$T$ denotes the temperature.

$P$ denotes the pressure.

Answer to Problem 6.64QE

Volume of hydrogen gas produced when $3.43\text{ g}$ of $\text{Fe}$ metal reacts with $40.0\text{ mL}$ of $2.43\text{ M HCl}$ is $0.5264\text{ L}$.

Explanation of Solution

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

  $\text{1 L}=\text{1000 mL}$

Hence convert $\text{40 mL}$ to $\text{mL}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{40 mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =0.04\text{ L}\end{array}$

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

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$        (1)

Substitute $\text{23 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (1).

  $\begin{array}{c}T\left(\text{K}\right)=\text{23}°\text{C}+273\text{ K}\\ =296\text{ K}\end{array}$

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

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

Substitute $\text{3}\text{.43 g}$ for mass and $\text{55}\text{.84 g/mol}$ for molar mass to calculate the number of moles of $\text{Fe}$ in equation (2).

  $\begin{array}{c}\text{Number of moles}=\frac{\text{3}\text{.43 g}}{\text{55}\text{.84 g/mol}}\\ =0.06142\text{ mol}\end{array}$

The formula to calculate molarity is given as follows:

  $\text{Molarity}\left(\text{mol/L}\right)=\frac{\text{number of moles}}{\text{volume}\left(\text{L}\right)}$        (3)

Substitute $\text{2}\text{.43 mol/L}$ for molarity and $\text{0}\text{.04 L}$ for volume in equation (3).

  $\text{2}\text{.43 mol/L}=\frac{\text{number of moles}}{\text{0}\text{.04 L}}$

Rearrange to calculate the number of moles of $\text{HCl}$.

  $\begin{array}{c}\text{Number of moles}=\left(\text{2}\text{.43 mol/L}\right)\text{0}\text{.04 L}\\ =\text{0}\text{.0972 mol}\end{array}$

Now the amount of hydrogen gas formed depends upon the limiting reactant. To determine the limiting reactants calculate the number of moles of ${\text{H}}_2$ produced from both the reactants the balanced chemical reaction when iron combines with $\text{HCl}$ is given as follows:

  $\text{Fe}+2\text{HCl}\to {\text{FeCl}}_2+{\text{H}}_2$        (4)

According to the stoichiometry of the balanced equation, one mole of iron metal produces one mole of ${\text{H}}_2$, therefore number of moles of ${\text{H}}_2$ produced from $0.06142\text{ mol Fe}$ is calculated as follows:

  $\begin{array}{c}{\text{Number of moles of H}}_2=\left(0.06142\text{ mol Fe}\right)\left(\frac{1{\text{ mol H}}_2}{\text{1 mol Fe}}\right)\\ =0.06142{\text{ mol H}}_2\end{array}$

Similarly, number of moles of ${\text{H}}_2$ produced from $\text{0}\text{.0972 mol}$ of $\text{HCl}$ is calculated as follows:

  $\begin{array}{c}{\text{Number of moles of H}}_2=\left(\text{0}\text{.0972 mol HCl}\right)\times \left(\frac{1{\text{ mol H}}_2}{\text{2 mol HCl }}\right)\\ =\text{0}{\text{.0486 mol H}}_2\end{array}$

Since $\text{HCl}$ yields less number of moles so it must be the limiting reagent, hence the number of moles of ${\text{H}}_2$ obtained from this reaction is $\text{0}\text{.0486 mol}$.

The formula to calculate volume as per the ideal gas equation is as follows:

  $V=\frac{nRT}{P}$        (5)

Substitute $\text{0}\text{.0486 mol}$ for $n$, $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$, $296\text{ K}$ for $T$ and $\text{2}\text{.25 atm}$ for $P$ in equation (5).

  $\begin{array}{c}V=\frac{\left(\text{0}\text{.0486 mol}\right)\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(296\text{ K}\right)}{\left(\text{2}\text{.25 atm}\right)}\\ =0.5246\text{ L}\end{array}$