Question

The production capacity for acrylonitrile (C3H3N) in the United States exceeds 2 million poundsper year. Acrylonitrile, the building block for polyacrylonitrilefibers and a variety of plastics, isproduced from gaseous propylene, ammonia, and oxygen:

$\mathbf{2}{\mathbf{\text{C}}}_{\mathbf{3}}{\mathbf{\text{H}}}_{\mathbf{6}}\left(\text{g}\right)\mathbf{+}\mathbf{2}{\mathbf{\text{NH}}}_{\mathbf{3}}\left(\text{g}\right)\mathbf{+}\mathbf{3}{\mathbf{\text{O}}}_{\mathbf{2}}\left(\text{g}\right)\mathbf{\to }\mathbf{2}{\mathbf{\text{C}}}_{\mathbf{3}}{\mathbf{\text{H}}}_{\mathbf{3}}\mathbf{\text{N}}\left(\text{g}\right)\mathbf{+}\mathbf{6}{\mathbf{\text{H}}}_{\mathbf{3}}\mathbf{\text{O}}\left(\text{g}\right)$

a. What mass of acrylonitrile can be produced from a mixture of 1.00 kg of propylene ${\mathbf{\text{C}}}_{\mathbf{3}}{\mathbf{\text{H}}}_{\mathbf{6}}$1.50 kg of ammonia, and 2.00 kg of oxygen, assuming 100% yield?
b. What mass of water is produced, and what masses of which starting materials are left inexcess?

Short answer

1. The mass of acrylonitrile that can be produced from a mixture of 1.00 kg of propylene${\text{C}}_3{\text{H}}_6$,1.50 kg of ammonia, and 2.00 kg of oxygen is$1.26\text{kg}$
2. The mass of water produced is $\text{1.28kg}$, and masses of starting materials left inexcess are 1.09 kg of${\mathbf{\text{NH}}}_{\mathbf{3}}$ is $\mathbf{1}\mathbf{.}\mathbf{09}\mathbf{}\mathbf{\text{kg}}$and 0.86 kg of${\mathbf{\text{O}}}_{\mathbf{2}}$ is $\mathbf{0}\mathbf{.}\mathbf{86}\mathbf{}\mathbf{\text{kg}}$.

Step-by-step solution

Definition

The mass of any material expressed in atomic mass units is quantitatively equivalent to its molar mass in grams per mole.

Determination of molar mass of acrylonitrile 

a)

The molar mass of ${\mathbf{\text{C}}}_{\mathbf{3}}{\mathbf{\text{H}}}_{\mathbf{6}}$:

$$\begin{gathered} =\text{12.01}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\text{×3+1.01}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol×6}$}\right. \\ \text{=42.09}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right. \end{gathered}$$

The molar mass of ${\mathbf{\text{NH}}}_3$:

$$\begin{gathered} 14.01\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.+1.01\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\times 3 \\ =17.09\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right. \end{gathered}$$

The molar mass of ${\mathbf{\text{O}}}_{\mathbf{2}}$:

$$\begin{gathered} 16.00\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\times 2 \\ =32.00\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right. \end{gathered}$$

The molar mass of ${\mathbf{\text{C}}}_{\mathbf{3}}{\mathbf{\text{H}}}_{\mathbf{3}}\mathbf{\text{N}}$:

$$\begin{gathered} =12.01\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\times 3+1.01\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\times 3+14.01\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right. \\ =53.07\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right. \end{gathered}$$

Determination of mass of acrylonitrile. 

The molar mass of${\text{C}}_3{\text{H}}_3\text{N}$from 1000g of${\text{C}}_3{\text{H}}_6$

$=\text{1.26g}$

The molar mass of ${\text{C}}_3{\text{H}}_3\text{N}$from 1500g of${\text{NH}}_3$

$=\text{4.56kg}$

The molar mass of${\text{C}}_3{\text{H}}_3\text{N}$from 2000g of${\text{O}}_2$:

$=2.21\text{g}$

role="math" localid="1648710671968" $2.21\text{kg}$is the smallest value, and the mass of acrylonitrile is role="math" localid="1648710641537" $\text{1.26kg}$ .

Determination of molar mass H2O 

b)

We get that is the limiting reactant

The molar mass of ${\text{H}}_2\text{O}$:

$\text{1.01}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\text{×2+16.00}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\text{=18.02}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

Let’s use ${\text{C}}_3{\text{H}}_6$ to determine themassof ${\text{H}}_2\text{O}$,

${\text{1000gC}}_3{\text{H}}_6\text{×}\frac{{\text{6molH}}_2\text{O}}{\text{42.09g}}\text{×}\frac{{\text{6molH}}_2\text{O}}{{\text{2molC}}_2{\text{H}}_6}\text{×}\frac{18\text{.02g}}{\text{1mol}}$

$\text{=1.28kg}$

Here,${\text{NH}}_3$ and ${\text{O}}_2$are the excess

Let’s calculate left over's

${\text{NH}}_3{\text{:1000gC}}_3{\text{H}}_6\text{×}\frac{{\text{1molC}}_3{\text{H}}_6}{42.09\text{g}}\text{×}\frac{{\text{2molNH}}_3}{{\text{2molC}}_3{\text{H}}_6}\text{×}\frac{17.04\text{g}}{1\text{mol}}$

$=0.14\text{g}$${\text{O}}_2{\text{:1000gC}}_2{\text{H}}_6\text{×}\frac{{\text{1molC}}_2{\text{H}}_6}{42.09\text{g}}\text{×}\frac{{\text{3molO}}_2}{{\text{2molC}}_2{\text{H}}_6}\text{×}\frac{32.00\text{g}}{1\text{mol}}$

$=\text{1.14g}$

Let’s subtract original mass to used mass

$$\begin{gathered} {\text{NH}}_3\text{=1.09kg} \\ {\text{O}}_{2=}\text{0.86kg} \end{gathered}$$

Therefore, the leftover masses are 1.09 kg of ${\mathbf{\text{NH}}}_{\mathbf{3}}$ is and 0.86 kg of ${\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}$