Chapter 5, Problem 68E

(a)

Interpretation Introduction

Interpretation:Percent yield of ${\text{HSiCl}}_{\text{3}}$ 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 68E

Percent yield of ${\text{HSiCl}}_{\text{3}}$ is $78.017\text{ \%}$ .

Explanation of Solution

Given Information:

Density of ${\text{HSiCl}}_{\text{3}}$ is $\text{0}\text{.850 g/mL}$ .

Pressure of $\text{HCl}$ is $\text{10 atm}$ .

Volume of $\text{HCl}$ is $\text{15}\text{.0 L}$ .

Temperature in Celsius is $\text{35 }°\text{C}$ .

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 $\text{35 }°\text{C}$ .

Substitute the value in the above formula.

  $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}\\ =\text{35 }°\text{C}+273\text{ K}\\ =308\text{ K}\end{array}$

The formula to calculate mass from density is as follows:

  $\text{Mass}=\left(\text{Volume}\right)\left(\text{Density}\right)$

For ${\text{HSiCl}}_{\text{3}}$

Volume is $\text{156 mL}$ .

Density is $\text{1}\text{.34 g/mL}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Mass}=\left(\text{Volume}\right)\left(\text{Density}\right)\\ =\left(\text{156 mL}\right)\left(\text{1}\text{.34 g/mL}\right)\\ =209.04\text{ g}\end{array}$

The expression to determine number of moles is as follows:

  $\text{Number of moles}=\frac{{\text{Mass of HSiCl}}_{\text{3}}}{{\text{Molar mass of HSiCl}}_{\text{3}}}$

Mass of ${\text{HSiCl}}_{\text{3}}$ is $209.04\text{ g}$ .

Molar mass of ${\text{HSiCl}}_{\text{3}}$ is $135.45\text{ g/mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of moles}=\frac{{\text{Mass of HSiCl}}_{\text{3}}}{{\text{Molar mass of HSiCl}}_{\text{3}}}\\ =\frac{209.04\text{ g}}{135.45\text{ g/mol}}\\ =1.3264\text{ mol}\end{array}$

The balanced equation for formation of ${\text{HSiCl}}_{\text{3}}$ is as follows:

  $\text{Si}\left(s\right)+3\text{HCl}\left(g\right)\to {\text{HSiCl}}_{\text{3}}+{\text{H}}_{\text{2}}\left(g\right)$

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

  $n=\frac{PV}{RT}$

The value of $P$ is $\text{10 atm}$ .

The value of $V$ is $\text{15}\text{.0 L}$ .

The value of $T$ is $308\text{ K}$ .

The value of $R$ is $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ .

Substitute the value in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{\left(\text{10 atm}\right)\left(\text{15}\text{.0 L}\right)}{\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(308\text{ K}\right)}\\ =5.9348\text{ mol}\end{array}$

In accordance with balanced equation, $\text{3 mol HCl}$ forms ${\text{1 mol HSiCl}}_{\text{3}}$ , thus moles of ${\text{HSiCl}}_{\text{3}}$ formed from $5.9348\text{ mol HCl}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of HSiCl}}_{\text{3}}=\left(5.9348\text{ mol HCl}\right)\left(\frac{{\text{1 mol HSiCl}}_{\text{3}}}{\text{3 mol HCl}}\right)\\ =1.9782{\text{ mol HSiCl}}_{\text{3}}\end{array}$

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

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

Number of moles is $1.9782\text{ mol}$ .

Molar mass is $135.45\text{ g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)\\ =\left(1.9782\text{ mol}\right)\left(135.45\text{ g/mol}\right)\\ =267.94\text{ g}\end{array}$

Thus, theoretical yield of ${\text{HSiCl}}_{\text{3}}$ expected $267.94\text{ g}$ .

The formula to calculate percent yield is as follows:

  $\text{Percent yield}=\left(\frac{\text{Observed yield}}{\text{Theoretical yield}}\right)\left(100\text{ \%}\right)$

Observed yield is $209.04\text{ g}$ .

Theoretical yield is $267.94\text{ g}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Percent yield}=\left(\frac{\text{Observed yield}}{\text{Theoretical yield}}\right)\left(100\text{ \%}\right)\\ =\left(\frac{209.04\text{ g}}{267.94\text{ g}}\right)\left(100\text{ \%}\right)\\ =78.017\text{ \%}\end{array}$

(b)

Interpretation Introduction

Interpretation:Volume of ${\text{SiH}}_4$ when percent yield of ${\text{HSiCl}}_{\text{3}}$ is $93.1\text{ \%}$ isto 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 68E

Volume of ${\text{SiH}}_4$ is $907.87\text{ mL}$ .

Explanation of Solution

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 $\text{35 }°\text{C}$ .

Substitute the value in the above formula.

  $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}\\ =\text{35 }°\text{C}+273\text{ K}\\ =308\text{ K}\end{array}$

The formula to calculate mass from density is as follows:

  $\text{Mass}=\left(\text{Volume}\right)\left(\text{Density}\right)$

For ${\text{HSiCl}}_{\text{3}}$

Volume is $\text{156 mL}$ .

Density is $\text{1}\text{.34 g/mL}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Mass}=\left(\text{Volume}\right)\left(\text{Density}\right)\\ =\left(\text{156 mL}\right)\left(\text{1}\text{.34 g/mL}\right)\\ =209.04\text{ g}\end{array}$

The expression to determine number of moles is as follows:

  $\text{Number of moles}=\frac{{\text{Mass of HSiCl}}_{\text{3}}}{{\text{Molar mass of HSiCl}}_{\text{3}}}$

Mass of ${\text{HSiCl}}_{\text{3}}$ is $209.04\text{ g}$ .

Molar mass of ${\text{HSiCl}}_{\text{3}}$ is $135.45\text{ g/mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of moles}=\frac{{\text{Mass of HSiCl}}_{\text{3}}}{{\text{Molar mass of HSiCl}}_{\text{3}}}\\ =\frac{209.04\text{ g}}{135.45\text{ g/mol}}\\ =1.5433\text{ mol}\end{array}$

The balanced equation for formation of ${\text{HSiCl}}_{\text{3}}$ is as follows:

  ${\text{4HSiCl}}_{\text{3}}\left(l\right)\to {\text{SiH}}_4+3{\text{SiCl}}_4\left(l\right)$

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

  $n=\frac{PV}{RT}$

The value of $P$ is $\text{10 atm}$ .

The value of $V$ is $\text{15}\text{.0 L}$ .

The value of $T$ is $308\text{ K}$ .

The value of $R$ is $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ .

Substitute the value in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{\left(\text{10 atm}\right)\left(\text{15}\text{.0 L}\right)}{\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(308\text{ K}\right)}\\ =5.9348\text{ mol}\end{array}$

In accordance with balanced equation, $4{\text{ mol HSiCl}}_{\text{3}}$ forms ${\text{1 mol SiH}}_4$ , thus moles of ${\text{SiH}}_4$ formed from $1.5433{\text{ mol HSiCl}}_{\text{3}}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of SiH}}_4=\left(1.5433{\text{ mol HSiCl}}_{\text{3}}\right)\left(\frac{{\text{1 mol SiH}}_4}{4{\text{ mol HSiCl}}_{\text{3}}}\right)\\ =0.385825{\text{ mol SiH}}_4\end{array}$

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

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

Number of moles is $0.385825\text{ mol}$ .

Molar mass is $\text{32}\text{.12 g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Mass}=\left(0.385825\text{ mol}\right)\left(\text{32}\text{.12 g/mol}\right)\\ =12.3929\text{ g}\end{array}$

The formula to calculate percent yield is as follows:

  $\text{Percent yield}=\left(\frac{\text{Observed yield}}{\text{Theoretical yield}}\right)\left(100\text{ \%}\right)$

Percent yield is $93.1\text{ \%}$ .

Theoretical yield is $12.3929\text{ g}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Percent yield}=\left(\frac{\text{Observed yield}}{\text{Theoretical yield}}\right)\left(100\text{ \%}\right)\\ 93.1\text{ \%}=\left(\frac{\text{Observed yield}}{12.3929\text{ g}}\right)\left(100\text{ \%}\right)\end{array}$

Rearrange to obtain the observed yield.

  $\begin{array}{c}\text{Observed yield}=\left(\frac{93.1\text{ \%}}{100\text{ \%}}\right)\left(12.3929\text{ g}\right)\\ =11.5377\text{ g}\end{array}$

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

  $V=\frac{mRT}{MP}$

The value of $P$ is $10\text{ atm}$ .

The value of $m$ is $11.5377\text{ g}$ .

The value of $M$ is $\text{32}\text{.12 g/mol}$ .

The value of $T$ is $308\text{ K}$ .

The value of $R$ is $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ .

Substitute the values in above equation.

  $\begin{array}{c}V=\frac{mRT}{MP}\\ =\frac{\left(11.5377\text{ g}\right)\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(308\text{ K}\right)}{\left(\text{32}\text{.12 g/mol}\right)\left(10\text{ atm}\right)}\\ =0.90787\text{ L}\end{array}$

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

  $\text{1 L}=\text{1000 mL}$

Hence convert $0.90787\text{ L}$ to $\text{mL}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(0.90787\text{ L}\right)\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =907.87\text{ mL}\end{array}$