Question 7.14: When silicon dioxide (sand) and carbon are heated, the produ...
When silicon dioxide (sand) and carbon are heated, the products are silicon carbide, SiC, and carbon monoxide. Silicon carbide is a ceramic material that tolerates extreme temperatures and is used as an abrasive and in the brake discs of cars. How many grams of CO are formed from 70.0 g of SiO_{2} and 50.0 g of C?
SiO_{2}(s) + 3C(s) \xrightarrow[]{\Delta } SiC(s) + 2CO(g)STEP 1 State the given and needed quantities (moles).
| ANALYZE THE PROBLEM | Given | Need | Connect |
| 70.0 g of SiO_{2}, 50.0 g of C | grams of CO from limiting reactant | molar masses, mole-mole factors | |
| Equation | |||
| SiO_{2}(s) + 3C(s) \xrightarrow[]{\Delta } SiC(s) + 2CO(g) | |||
STEP 2 Write a plan to convert the quantity (grams) of each reactant to quantity (grams) of product.
grams of SiO_{2} \boxed{\begin{array}{l}\text{Molar}\\\text{Mass}\end{array}} → moles of SiO_{2} \boxed{\begin{array}{l}\text{Mole-mole}\\\text{factor}\end{array}} → moles of CO \boxed{\begin{array}{l}\text{Molar}\\\text{Mass}\end{array}} → grams of COgrams of C \boxed{\begin{array}{l}\text{Molar}\\\text{Mass}\end{array}} → moles of C \boxed{\begin{array}{l}\text{Mole-mole}\\\text{factor}\end{array}} → moles of CO \boxed{\begin{array}{l}\text{Molar}\\\text{Mass}\end{array}} → grams of CO
STEP 3 Use coefficients to write mole-mole factors.
\begin{array}{r c}\boxed{\begin{matrix} 1 mole of SiO_{2} = 60.09 g of SiO_{2} \\\frac{ 60.09 g SiO_{2}}{1 mole SiO_{2}} \text{ and } \frac{1 mole SiO_{2}}{ 60.09 g SiO_{2}} \end{matrix}} & \boxed{\begin{matrix}1 mole of C = 12.01 g of C\\ \frac{ 12.01 g C}{ 1 mole C}\text{ and }\frac{ 1 mole C}{ 12.01 g C} \end{matrix}} & \boxed{\begin{matrix}1 mole of CO = 28.01 g of CO\\ \frac{ 28.01 g CO}{ 1 mole CO}\text{ and }\frac{ 1 mole CO}{ 28.01 g C} \end{matrix}}\end{array}\begin{array}{r c}\boxed{\begin{matrix} 2 moles of CO = 1 mole of SiO_{2} \\\frac{ 2 moles CO}{1 mole SiO_{2}} \text{ and } \frac{1 mole SiO_{2}}{ 2 moles CO} \end{matrix}} & \boxed{\begin{matrix}2 moles of CO = 3 moles of C\\ \frac{ 2 moles CO}{ 3 moles C}\text{ and }\frac{ 3 moles C}{ 2 moles CO} \end{matrix}}\end{array}
STEP 4 Calculate the quantity (moles) of product from each reactant, and select the smaller quantity (moles) as the limiting reactant.
Grams of CO (product) from SiO_{2}:
Grams of CO (product) from C:
\overset{Three SFs}{50.0 \cancel{g C}} \times \overset{Exact}{\underset{Four SFs}{\boxed{\frac{1 \cancel{mole C}}{12.01 \cancel{ g C}} }}} \times \overset{Exact}{\underset{Exact}{\boxed{\frac{2 \cancel{moles CO}}{3 \cancel{ moles CO}} }}} \times \overset{Four SFs}{\underset{Exact}{\boxed{\frac{28.01 g CO}{1 \cancel{ mole CO}} }}} = \overset{Three SFs}{77.7 g of CO}The smaller amount, 65.3 g of CO, is the most CO that can be produced. This also means that the SiO_{2} is the limiting reactant.