Q14.147CP - Chemistry: Molecular Nature Of Matter And Change

Question

Semiconductors made from elements in Groups 3A(13) and 5A(15) are typically prepared by direct reaction of the elements at high temperature. An engineer treats 32.5 g of molten gallium with 20.4 L of white phosphorus vapor at 515 K and 195 kPa. If purification losses are 7.2% by mass, how many grams of gallium phosphide will be prepared?

Step-by-Step Solution

Answer

43.54 g GaP gallium phosphide will be produced. 

Step 1: Definition of Chemical reaction

Chemical equation that is balanced:

A balanced chemical equation is one in which each side of the reaction has the same amount of atoms and elements.

The following is the balanced chemical reaction between molten gallium and white phosphorus:

${\text{4Ga}}_{\text{(l)}}{\text{ + P}}_{\text{4(g)}}\to {\text{4GaP}}_{\text{(s)}}$

To begin, determine the moles of gallium and white phosphorus as follows:

Gallium moles are computed using its molar mass as follows:

$\begin{array}{rcl}\mathrm{The} \mathrm{moles} \mathrm{of} \mathrm{Ga}& =& \frac{\mathrm{Amount} \mathrm{of} \mathrm{Ga}}{\mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{Ga}}\\ & =& \frac{32.5 \mathrm{g}}{69.7 \mathrm{g}/\mathrm{mol}}\\ & =& 0.466 \mathrm{mol}\\ & & \end{array}$

The moles of white phosphorus will be calculated as follows:

The ideal gas law is following:

PV = nRT

Here,  $\mathrm{R};\left(0.0821\mathrm{atm}.{\mathrm{Lmol}}^{-1}.{\mathrm{K}}^{-1}\right)$ is the universal gas constant.

To calculate the number of moles of white phosphorus at 195 kPa and 515 K, uses the following expression:

$\begin{array}{rcl}PV& =& nRT\\ n& =& \frac{PV}{RT}\\ & & \end{array}$

Here, n is the number of moles, p is the pressure, v is the volume = 20.4-mL, and T is the temperature.

First, change the unit of pressure from kPa to atm as follows:

$$\begin{gathered} 1.0 \mathrm{atm}=101.33\mathrm{kPa} \\ 195\mathrm{kPa}\times \frac{1.0\mathrm{atm}}{101.33\mathrm{kPa}}=1.92\mathrm{atm} \end{gathered}$$

Now calculate the number of moles by putting all the values in the above ideal gas law:

$$\begin{gathered} \mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}} \\ \mathrm{n}=\frac{1.92 \mathrm{atm}\times 20.4 \mathrm{L}}{0.8206 \mathrm{atm}.\mathrm{Lmol}-1.\mathrm{K}-1\times 515 \mathrm{K}} \\ =0.927 \mathrm{mol} \end{gathered}$$

Step 2: Calculation of moles

Now calculate the moles of GaP by using the moles of white phosphorus and gallium:

$4{\mathrm{Ga}}_{\left(\mathrm{l}\right)}+{\mathrm{P}}_{4\left(\mathrm{g}\right)}\to 4{\mathrm{GaP}}_{\left(\mathrm{s}\right)}$

$\begin{array}{rcl}\mathrm{Moles} \mathrm{of} \mathrm{GaP}& =& \frac{4 \mathrm{Mol} \mathrm{GaP}}{4 \mathrm{Mol} \mathrm{Ga}}\times 0.466 \mathrm{mol} \mathrm{Ga}\\ & =& 0.466 \mathrm{Mol} \mathrm{GaP}\\ \mathrm{Mol} \mathrm{of} \mathrm{GaP}& =& \frac{4 \mathrm{Mol} \mathrm{GaP}}{1 \mathrm{Mol} {\mathrm{P}}_4}\times 0.927 \mathrm{mol} {\mathrm{P}}_4\\ & =& 3.708 \mathrm{Mol} \mathrm{GaP}\end{array}$

Therefore lesser number of products will obtained with Ga, hence is the limiting reactant.

Step 3: Calculation of mass of products

Now calculated the mass of products by their molar mass as follows:

$\begin{array}{rcl}\mathrm{Mass} \mathrm{of} \mathrm{GaP}& =& \frac{100.697 \mathrm{g} \mathrm{GaP}}{1 \mathrm{Mole} \mathrm{GaP}}\times 0.466 \mathrm{GaP} \mathrm{Mole}\\ & =& 46.92\mathrm{g} \mathrm{GaP}\end{array}$

Step 4: Amount of GaP production

This 46.92 g GaP amount of GaP is the 100% yield. To determine the actual yield of GaP minus the amount of GaP which lost in the purification process.

100%-7.2% = 92.8%

Now calculate the actual amount of, which is 92.8 %.

$\begin{array}{rcl}46.92 \mathrm{g} \mathrm{GaP}\times 92.8\%& =& 46.92 \mathrm{g} \mathrm{GaP}\times \frac{92.8}{100}\\ & =& 43.54 \mathrm{g} \mathrm{GaP}\end{array}$

 

Hence 43.54 g GaP gallium phosphide will be produced.