Chapter 4, Problem 4.49PAE

4.47 Silicon carbide is, an abrasive used in the manufacture of grinding wheels. A company is investigating whether they can be more efficient in the construction of such wheels by making their own SiC via the reaction

   ${\text{SiO}}_{\text{2}}+\text{ 3 C }\to \text{ SiC }+\text{ 2 CO}$

(a) If this reaction consistently has a yield of 85%, what is the minimum amount of both silicon dioxide and carbon needed to produce 3400 kg of SiC for the manufacture of cutting wheels? (b) The silicon dioxide is to he obtained from sand and the carbon is derived from coal. If the available sand is 95% SiO₂ by weight and the coal is 73% C by weight, what mass of coal is needed for each metric ton of sand used?

Interpretation Introduction

To determine: The minimum amount of both silicon dioxide and carbon needed to produce $\text{34}00\text{ kg}$ of SiC for the manufacture of cutting wheels.

Explanation of Solution

Step 1: Balance Reaction

Reaction is already balanced

Step 2: Change all amounts to mol

$\text{mol}\text{SiC=}\frac{\text{mass}}{\text{Molecular}\text{Weight}}\text{=}\frac{\text{3400}}{\text{40}\text{.09}}\text{= 84}\text{.809 kmol}$

Step 3: Relate stoichiometry

$\text{Ratio=1:1}$

$\begin{array}{l}\text{1 mol }{\text{SiO}}_{\text{2}}\text{=1 }\text{mol }\text{SiC}\\ \text{mol }{\text{SiO}}_{\text{2}}\text{=84}\text{.809 kmol}\end{array}$

Now, for carbon:

$\text{Ratio=1:3}$

$\begin{array}{l}\text{1 mol}{\text{SiO}}_{\text{2}}\text{=3 mol}\text{C}\\ \text{mol}\text{C=3}\times \text{(84}\text{.809 kmol) = 254}\text{.427}\text{kmol}\text{C}\end{array}$

Step 4: Calculate mass

For SiO₂

$\begin{array}{l}\text{mass}{\text{SiO}}_{\text{2}}\text{=mol}\times {\text{Molecular Weight SiO}}_{\text{2}}\\ \text{mass}{\text{SiO}}_{\text{2}}\text{=(84}\text{.809}\text{kmol)}\times \text{(60}\text{.0843 kg/kmol)}\\ \text{mass}{\text{SiO}}_{\text{2}}\text{= 5095}\text{.689}\text{kg}{\text{SiO}}_{\text{2}}\end{array}$

Now, for carbon

$\begin{array}{l}\text{mass}\text{C=mol}\times {\text{Molecular Weight SiO}}_{\text{2}}\\ \text{mass}\text{C=(254}\text{.427 kmol)}\times \text{(12}\text{.0 kg/kmol)}\\ \text{mass}\text{C=3052}\text{.8}\text{kg}\text{C}\end{array}$

Step 5: Correct for actual yield

$\begin{array}{l}\text{actual}\text{mass}{\text{SiO}}_{\text{2}}\text{=}\left(\text{5095}\text{.689}\text{kg}{\text{SiO}}_{\text{2}}\right)\times \left(\frac{\text{85}\text{\%}}{\text{100}\text{\%}}\right)\\ \text{actual}\text{mass}{\text{SiO}}_{\text{2}}\text{=4330}\text{.75 }\text{kg}\end{array}$

$\begin{array}{l}\text{actual}\text{mass }\text{C=}\left(\text{3052}\text{.8 }\text{kg}\text{C}\right)\text{×}\left(\frac{\text{85 \%}}{\text{100 \%}}\right)\\ \text{actual}\text{mass }\text{C=2594}\text{.88 }\text{kg}\end{array}$

Interpretation Introduction

To determine:

Mass of coal needed for each metric ton of sand used.

Explanation of Solution

Now, relate new amounts.

Step 1: Relate 1 metric ton of sand:

$\begin{array}{l}\text{mass}\text{Sand=1}\text{ton=1000}\text{kg}\\ \text{mass}{\text{SiO}}_{\text{2}}\text{=(1000}\text{kg)}\times \left(\frac{\text{95}}{\text{100}}\right)\\ \text{mass}{\text{SiO}}_{\text{2}}\text{=950}\text{kg}\end{array}$

Step 2: Use Ratio between reactants:

$\begin{array}{l}\text{Ratio=}\frac{\text{3}\times \text{Molecular Weight }\text{C}}{\text{1}\times \text{Molecular Weight}{\text{SiO}}_{\text{2}}}\text{=}\frac{\text{3}\times \text{(12)}}{\text{60}\text{.08}}\text{=0}\text{.5992}\\ \text{mass}\text{C=}\left(\text{950}\text{kg}\right)\times \left(\text{0}\text{.5992}\right)\text{=569}\text{.24}\text{kg}\text{C}\end{array}$

Step 3:Calculate mass of coal

$\begin{array}{l}\text{mass}\text{Coal=}\frac{\text{569}\text{.24}\text{kg}\text{C}}{\text{1}}\times \frac{\text{100}\text{kg}\text{Coal}}{\text{73}\text{kg}\text{C}}\\ \text{mass}\text{Coal=779}\text{.78}\text{kg}\end{array}$

Conclusion

a. $\text{mass}\text{C=2594}\text{.88}\text{kg}$ ; $\text{mass}{\text{SiO}}_{\text{2}}\text{=4330}\text{.75 }\text{kg}$

b. $\text{mass}\text{Coal=779}\text{.78}\text{kg}\text{per ton}$