Chapter 5, Problem 162IP

Silane, SiH₄, is the silicon analogue of methane, CH4. It is prepared industrially according to the following equations:

$\begin{array}{l}\text{Si}\left(s\right)+\text{ 3HC1}\left(g\right)\to {\text{ HSiCl}}_{\text{3}}\left(l\right)+{\text{ H}}_{\text{2}}\left(g\right)\hfill \\ {\text{4HSiCl}}_{\text{3}}\left(l\right)\to {\text{ SiH}}_{\text{4}}\left(g\right)+{\text{ 3SiCl}}_{\text{4}}\left(l\right)\hfill \end{array}$

a. If 156 mL HSiCl₃ (d = 1.34 g/mL) is isolated when 15.0 L HC1 at 10.0 atm and 35°C is used, what is the percent yield of HSiCl₃?

b. When 156 mL HSiCl₃ is heated, what volume of SiH₄ at 10.0 atm and 35°C will be obtained if the percent yield of the reaction is 93.1%?

Interpretation Introduction

Interpretation: The percent yield of ${\text{HSiCl}}_{\text{3}}$ in the given reaction has to be calculated.

Concept introduction:

• Balance equation of a reaction is written according to law of conservation of mass.

• Number of moles of a substance, According to ideal gas equation,

$\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

• Equation for mass of a substance from its density and volume, 

  $\text{Mass}\text{=}\text{Density}\times \text{Volume}$

• Percent yield of reaction is the ratio of mass of actual yield to the mass of theoretical yield.

• Mole ratios between the reactant and a product of a reaction are depends upon the coefficients of reactant and product in a balanced chemical equation.

• Mass of a substance produced from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{produced}\text{mass}$

• Number of moles of a substance from its given mass is,

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

• According to ideal gas equation for finding volume of a gas,

$\text{Volume}\text{=}\frac{\text{Number}\text{of}\text{moles}\times \text{R}\times \text{Temperature}}{\text{pressure}}$

To determine: the percent yield of ${\text{HSiCl}}_{\text{3}}$ in the given reaction.

Answer to Problem 162IP

The percent yield of ${\text{HSiCl}}_{\text{3}}$ is $78\text{\%}$.

Explanation of Solution

To find: the number of moles of $\text{HCl}$ reacted in the given reaction.

The number of moles of $\text{HCl}$ is $\text{5}\text{.93}\text{mol}$.

The volume of $\text{HCl}$ in the given reaction is given as $\text{15}\text{L}$.

The pressure is given as $\text{10}\text{atm}$.

The temperature is given as ${\text{35}}^{\text{o}}\text{C}\text{=}\text{(35+273)}\text{K}\text{=}308\text{K}$.

The equation for finding number of moles of a substance, According to ideal gas equation,

$\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

Therefore,

The number of moles of $\text{HCl}$ is,

${\text{n}}_{\text{HCl}}\text{=}\frac{\text{15L}\text{×}\text{10}\text{atm}}{\text{0}\text{.08206}\text{L}\text{atm/Kmol}\text{×}308\text{K}}\text{=}\text{5}\text{.93}\text{mol}$

Hence,

The number of moles of $\text{HCl}$ is $\text{5}\text{.93}\text{mol}$.

To find: the number of moles ${\text{HSiCl}}_{\text{3}}$ in the given reaction.

The number of moles of ${\text{HSiCl}}_{\text{3}}$ in the given reaction is $\text{1}\text{.976}\text{mol}$.

The balanced chemical equation for the given reaction is,

$\text{Si}\text{+}3\text{HCl}\to {\text{HSiCl}}_{\text{3}}\text{+}{\text{H}}_{\text{2}}$

Here,

The mole ratio between ${\text{HSiCl}}_{\text{3}}$ and $\text{HCl}$ is $\text{1:}3$.

The number of moles of $\text{HCl}$ is $\text{5}\text{.93}\text{mol}$, and then the number of moles ${\text{HSiCl}}_{\text{3}}$ is,

$\text{5}\text{.93}\text{mol}\text{HCl}\text{×}\frac{1\text{mol}{\text{HSiCl}}_{\text{3}}}{3\text{mol}\text{HCl}}\text{=}\text{1}\text{.976}\text{mol}{\text{HSiCl}}_{\text{3}}$

Therefore,

Theoretical calculated number of moles of ${\text{HSiCl}}_{\text{3}}$ in the given reaction is $\text{1}\text{.976}\text{mol}$.

To determine: the percent yield of ${\text{HSiCl}}_{\text{3}}$ in the given reaction.

The percent yield of ${\text{HSiCl}}_{\text{3}}$ is $78\text{\%}$.

• The volume of obtained ${\text{HSiCl}}_{\text{3}}$ is given as $\text{156}\text{mL}\text{=}\text{0}\text{.156}\text{L}$.

• The density of obtained ${\text{HSiCl}}_{\text{3}}$ is given as $\text{1}\text{.34g/L}$.

  Equation for mass of a substance from its density and volume,

$\text{Mass}\text{=}\text{Density}\times \text{Volume}$

Therefore,

The mass obtained ${\text{HSiCl}}_{\text{3}}$ is,

$\text{1}\text{.34g/L}\text{×}\text{0}\text{.156}\text{L}\text{=}\text{209}\text{g}$

Hence,

The actual yield of ${\text{HSiCl}}_{\text{3}}$ in the given reaction is $\text{209}\text{g}$.

• Theoretical calculated number of moles of ${\text{HSiCl}}_{\text{3}}$ in the given reaction is $\text{1}\text{.976}\text{mol}$.

  Mass of a substance produced from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{produced}\text{mass}$

Therefore,

The mass of ${\text{HSiCl}}_{\text{3}}$ in the given reaction is,

$\text{1}\text{.976}\text{mol}\text{×}\text{135}\text{.45}\text{g}\text{=}\text{268}\text{g}$

Hence,

The theoretical yield of ${\text{HSiCl}}_{\text{3}}$ in the given reaction is $\text{268}\text{g}$.

• Percent yield of reaction is the ratio of mass of actual yield to the mass of theoretical yield.

  The actual yield of ${\text{HSiCl}}_{\text{3}}$ in the given reaction is $\text{209}\text{g}$.

  The theoretical yield of ${\text{HSiCl}}_{\text{3}}$ in the given reaction is $\text{268}\text{g}$.

  Therefore,

  The percent yield of ${\text{HSiCl}}_{\text{3}}$ is,

  $\frac{\text{209}\text{g}}{\text{268}\text{g}}=78\text{\%}$

Hence,

The percent yield of ${\text{HSiCl}}_{\text{3}}$ is $78\text{\%}$.

Interpretation Introduction

Interpretation: The volume of ${\text{SiH}}_4$ obtained in the given reaction has to be calculated.

Concept introduction:

• Balance equation of a reaction is written according to law of conservation of mass.

• Number of moles of a substance, According to ideal gas equation,

$\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

• Equation for mass of a substance from its density and volume, 

  $\text{Mass}\text{=}\text{Density}\times \text{Volume}$

• Percent yield of reaction is the ratio of mass of actual yield to the mass of theoretical yield.

• Mole ratios between the reactant and a product of a reaction are depends upon the coefficients of reactant and product in a balanced chemical equation.

• Mass of a substance produced from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{produced}\text{mass}$

• Number of moles of a substance from its given mass is,

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

• According to ideal gas equation for finding volume of a gas,

$\text{Volume}\text{=}\frac{\text{Number}\text{of}\text{moles}\times \text{R}\times \text{Temperature}}{\text{pressure}}$

To determine: the percent yield of ${\text{HSiCl}}_{\text{3}}$ and the volume of ${\text{SiH}}_4$ obtained in the given reaction.

Answer to Problem 162IP

The volume of ${\text{SiH}}_4$ obtained in the given reaction is $\text{0}\text{.907L}$.

Explanation of Solution

To find: the number of moles of ${\text{SiH}}_4$ in the given reaction $\text{4}{\text{HSiCl}}_3\to {\text{SiH}}_4\text{+}3{\text{SiCl}}_{\text{4}}$.

The number of moles of ${\text{SiH}}_4$ in the given reaction is $\text{0}\text{.386}\text{mol}$.

• The actual yield of ${\text{HSiCl}}_{\text{3}}$ in the given reaction is $\text{209}\text{g}$.

  Equation of Number of moles of a substance from its given mass is,

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

Therefore,

The number of moles of ${\text{HSiCl}}_{\text{3}}$ in the given reaction is,

$\text{Number of moles}\text{=}\frac{\text{209}\text{g}}{\text{135}\text{.45}\text{g}}=1.543\text{mol}$

• The balanced chemical equation for the given reaction is,

  $\text{4}{\text{HSiCl}}_3\to {\text{SiH}}_4\text{+}3{\text{SiCl}}_{\text{4}}$

  Here,

  The mole ratio between ${\text{SiH}}_4$ and ${\text{HSiCl}}_{\text{3}}$ is $\text{1:}4$.

  The number of moles ${\text{HSiCl}}_{\text{3}}$ is calculated as $1.543\text{mol}$, and then the number of moles ${\text{SiH}}_4$ is,

$1.543\text{mol}{\text{HSiCl}}_{\text{3}}\text{×}\frac{1\text{mol}{\text{SiH}}_4}{4\text{mol}{\text{HSiCl}}_{\text{3}}}\text{=}\text{0}\text{.386}\text{mol}{\text{SiH}}_4$

Hence,

The theoretically calculated number of moles of ${\text{SiH}}_4$ in the given reaction is $\text{0}\text{.386}\text{mol}$.

To determine: the number of moles of ${\text{SiH}}_4$ obtained in the given reaction from its given percent yield.

The number of moles of ${\text{SiH}}_4$ obtained in the given reaction is $\text{0}\text{.359}\text{mol}$.

The Percent yield of ${\text{SiH}}_4$ in the given reaction is given as $93.1\%$.

Theoretically calculated number of moles of ${\text{SiH}}_4$ in the given reaction is $\text{0}\text{.386}\text{mol}$.

Therefore,

The actual number of moles of ${\text{SiH}}_4$ in the given reaction is,

$\text{0}\text{.386}\text{mol}\times 0.931=0.359\text{mol}$

Hence,

The number of moles of ${\text{SiH}}_4$ obtained in the given reaction is $\text{0}\text{.359}\text{mol}$.

To determine: the volume of ${\text{SiH}}_4$ obtained in the given reaction $\text{4}{\text{HSiCl}}_3\to {\text{SiH}}_4\text{+}3{\text{SiCl}}_{\text{4}}$.

The volume of ${\text{SiH}}_4$ obtained in the given reaction is $\text{0}\text{.907L}$.

The number of moles of ${\text{SiH}}_4$ obtained in the given reaction is $\text{0}\text{.359}\text{mol}$.

The pressure is given as $10\text{atm}$.

The temperature is given as ${\text{35}}^{\text{o}}\text{C}\text{=}\text{(35+273)}\text{K}\text{=}\text{308}\text{K}$.

The equation for finding volume of a gas is,

According to ideal gas equation,

$\text{Volume}\text{=}\frac{\text{Number}\text{of}\text{moles}\times \text{R}\times \text{Temperature}}{\text{pressure}}$

Therefore,

$\begin{array}{l}\text{Volume}{\text{of SiH}}_4\text{=}\frac{\text{0}\text{.359}\text{mol}\times \text{0}\text{.08206}\text{L}\text{atm/K}\text{mol}\times 308\text{K}}{10\text{atm}}\\ =0.907\text{L}\end{array}$

Hence,

The volume of ${\text{SiH}}_4$ obtained in the given reaction is $\text{0}\text{.907L}$.