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 SiO2 and

Question

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 [latex]SiO_{2}[/latex] and 50.0 g of C?

[latex]SiO_{2}(s) + 3C(s) \xrightarrow[]{\Delta } SiC(s) + 2CO(g)[/latex]

Accepted Answer

STEP 1 State the given and needed quantities (moles).

ANALYZE THE PROBLEM	Given	Need	Connect
	70.0 g of [latex]SiO_{2}[/latex], 50.0 g of C	grams of CO from limiting reactant	molar masses, mole-mole factors
	Equation
	[latex] SiO_{2}(s) + 3C(s) \xrightarrow[]{\Delta } SiC(s) + 2CO(g) [/latex]

STEP 2 Write a plan to convert the quantity (grams) of each reactant to quantity (grams) of product.

[latex] grams of SiO_{2} \boxed{\begin{array}{l} ext{Molar}\ ext{Mass}\end{array}} → moles of SiO_{2} \boxed{\begin{array}{l} ext{Mole-mole}\ ext{factor}\end{array}} → moles of CO \boxed{\begin{array}{l} ext{Molar}\ ext{Mass}\end{array}} → grams of CO[/latex]

[latex] grams of C \boxed{\begin{array}{l} ext{Molar}\ ext{Mass}\end{array}} → moles of C \boxed{\begin{array}{l} ext{Mole-mole}\ ext{factor}\end{array}} → moles of CO \boxed{\begin{array}{l} ext{Molar}\ ext{Mass}\end{array}} → grams of CO[/latex]

STEP 3 Use coefficients to write mole-mole factors.

[latex] \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}} ext{ 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} ext{ 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} ext{ and }\frac{ 1 mole CO}{ 28.01 g C} \end{matrix}}\end{array} [/latex]

[latex] \begin{array}{r c}\boxed{\begin{matrix} 2 moles of CO = 1 mole of SiO_{2} \\frac{ 2 moles CO}{1 mole SiO_{2}} ext{ 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} ext{ and }\frac{ 3 moles C}{ 2 moles CO} \end{matrix}}\end{array} [/latex]

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 [latex]SiO_{2}[/latex]:

[latex] \overset{Three SFs}{\underset{Limiting reactant }{70.0 \cancel{g SiO_{2}}}} imes \overset{Exact}{\underset{Four SFs}{\boxed{\frac{1 \cancel{mole SiO_{2}}}{60.09 \cancel{ g SiO_{2}}} }}} imes \overset{Exact}{\underset{Exact}{\boxed{\frac{2 \cancel{mole CO}}{1 \cancel{ mole SiO_{2}}} }}} imes \overset{Four SFs}{\underset{Exact}{\boxed{\frac{28.01 g CO}{1 \cancel{ mole CO}} }}} = \overset{Three SFs}{\underset{\begin{array}{l} Smaller amount\ of product \end{array} }{65.3 g of CO}} [/latex]

Grams of CO (product) from C:

[latex] \overset{Three SFs}{50.0 \cancel{g C}} imes \overset{Exact}{\underset{Four SFs}{\boxed{\frac{1 \cancel{mole C}}{12.01 \cancel{ g C}} }}} imes \overset{Exact}{\underset{Exact}{\boxed{\frac{2 \cancel{moles CO}}{3 \cancel{ moles CO}} }}} imes \overset{Four SFs}{\underset{Exact}{\boxed{\frac{28.01 g CO}{1 \cancel{ mole CO}} }}} = \overset{Three SFs}{77.7 g of CO}[/latex]

The smaller amount, 65.3 g of CO, is the most CO that can be produced. This also means that the [latex]SiO_{2}[/latex] is the limiting reactant.

Question 7.14: When silicon dioxide (sand) and carbon are heated, the produ...

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