Chapter 5, Problem 61E

Interpretation Introduction

Interpretation: Mass of iron splints and $98\text{ \%}$ by mass ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ needed to completely fill balloon is to be calculated.

Concept introduction: The expression to determine number of moles is as follows:

  $\text{Number of moles}=\frac{\text{Given Mass}}{\text{Molar mass}}$

The formula to calculate moles as per ideal gas law is as follows:

  $n=\frac{PV}{RT}$

Where,

• $R$ is gas constant.
• $V$ denotes the volume.
• $n$ denotes number of moles.
• $T$ denotes temperature.
• $P$ denotes pressure.

Answer to Problem 61E

Mass of iron splints and $98\text{ \%}$ by mass ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ needed to completely fill balloonare $1.495\times {10}^7\text{ g}$ and $2.572\times {10}^7\text{ g}$ respectively.

Explanation of Solution

Given Information:

Mass percent of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $98\text{ \%}$ .

Pressure of balloon is $\text{1 atm}$ .

Loss of ${\text{H}}_2$ is $\text{20 \%}$ .

Temperature in Celsius is $\text{0 }°\text{C}$ .

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

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

Temperature in Celsius is $\text{0 }°\text{C}$ .

Substitute the value in the above formula.

  $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}\\ =0°\text{C}+273\text{ K}\\ =273\text{ K}\end{array}$

The formula to percent of ${\text{H}}_2$ is as follows:

  $\text{\% of}{\text{H}}_2\text{ inside}=\left(\frac{{\text{V of H}}_2\text{ in balloon}}{{\text{Total V of H}}_2\text{ in balloon}}\right)\left(100\right)$

Percent of ${\text{H}}_2$ inside is $\text{80 \%}$ .

  $\text{V}$ of ${\text{H}}_2$ is ${\text{4800 m}}^3$ .

Substitute the value in above formula to commute

  $\begin{array}{c}\text{\% of}{\text{H}}_2\text{ inside}=\left(\frac{{\text{V of H}}_2\text{ in balloon}}{{\text{Total V of H}}_2\text{ in balloon}}\right)\\ \text{80 \%}=\left(\frac{4800{\text{ m}}^3}{{\text{Total Vof H}}_2\text{ in balloon}}\right)\end{array}$

Rearrange to obtain value of total $\text{V}$ of ${\text{H}}_2$ in balloon.

  $\begin{array}{c}TotalVof{H}_{2}inballoon=\left(\frac{4800m^3}{0.80}\right)\\ =6000m^3\end{array}$

The conversion factor to convert ${\text{m}}^3$ to ${\text{dm}}^3$ is as follows:

  $1{\text{ m}}^3={10}^3{\text{ dm}}^3$

Thus, $6000m^3$ is converted to $\text{atm}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(6000m^3\right)\left(\frac{{10}^3{\text{ dm}}^3}{1{\text{ m}}^3}\right)\\ =6\times {10}^6{\text{ dm}}^3\end{array}$

The formula to calculate moles as per ideal gas law is as follows:

  $n=\frac{PV}{RT}$

The value of $P$ is $\text{1 atm}$ .

The value of $V$ is $6\times {10}^6{\text{ dm}}^3$ .

The value of $T$ is $\text{273 K}$ .

The value of $R$ is $0.0821{\text{ dm}}^3\cdot \text{atm/mol}\cdot \text{K}$ .

Substitute the value in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{\left(\text{1 atm}\right)\left(6\times {10}^6{\text{ dm}}^3\right)}{\left(0.08206{\text{ dm}}^3\cdot \text{atm/mol}\cdot \text{K}\right)\left(\text{273 K}\right)}\\ =2.6769\times {10}^5\text{ mol}\end{array}$

The balanced equation is given as follows:

  $\text{Fe}\left(s\right)+{\text{H}}_{\text{2}}{\text{SO}}_4\left(aq\right)\to {\text{FeSO}}_4\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

In accordance with above balanced equation, ${\text{1 mol H}}_2$ forms from $\text{1 mol Fe}$ , thus moles of $\text{Fe}$ formed from $2.6769\times {10}^5{\text{ mol H}}_2$ is calculated as follows:

  $\begin{array}{c}\text{Moles of Fe}=\left(2.6769\times {10}^5{\text{ mol H}}_2\right)\left(\frac{\text{1 mol Fe}}{{\text{1 mol H}}_2}\right)\\ =2.6769\times {10}^5\text{ mol Fe}\end{array}$

In accordance with balanced equation, ${\text{1 mol H}}_2$ formsfrom ${\text{1 mol H}}_2{\text{SO}}_{\text{4}}$ , thus moles of ${\text{H}}_2{\text{SO}}_{\text{4}}$ formed from $2.6769\times {10}^5{\text{ mol H}}_2$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of H}}_2{\text{SO}}_{\text{4}}=\left(2.6769\times {10}^5{\text{ mol H}}_2\right)\left(\frac{{\text{1 mol H}}_2{\text{SO}}_{\text{4}}}{{\text{1 mol H}}_2}\right)\\ =2.6769\times {10}^5{\text{ mol H}}_2{\text{SO}}_{\text{4}}\end{array}$

The formula to calculate mass from moles is given as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$

For $\text{Fe}$

Number of moles is $2.6769\times {10}^5\text{ mol}$ .

Molar mass is $\text{55}\text{.845 g/mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Mass}=\left(2.6769\times {10}^5\text{ mol}\right)\left(\text{55}\text{.845 g/mol}\right)\\ =1.495\times {10}^7\text{ g}\end{array}$

For ${\text{H}}_2{\text{SO}}_{\text{4}}$

Number of moles is $2.6769\times {10}^5\text{ mol}$ .

Molar mass is $\text{98}\text{.079 g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Mass}=\left(2.6769\times {10}^5\text{ mol}\right)\left(\text{98}\text{.079 g/mol}\right)\\ =2.625\times {10}^7\text{ g}\end{array}$

Since mass of ${\text{H}}_2{\text{SO}}_{\text{4}}$ is $98\text{ \%}$ , thus actual mass of ${\text{H}}_2{\text{SO}}_{\text{4}}$ is calculated as follows:

  $\begin{array}{c}{\text{Mass of H}}_2{\text{SO}}_{\text{4}}=\left(98\text{ \%}\right)\left(2.625\times {10}^7\text{ g}\right)\\ =2.572\times {10}^7\text{ g}\end{array}$

Hence, mass of $\text{Fe}$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ needed to completely fill balloon are $1.495\times {10}^7\text{ g}$ and $2.572\times {10}^7\text{ g}$ respectively.