Question 5.SP.19: When charcoal is burned, the carbon (C) it contains reacts w...
When charcoal is burned, the carbon (C) it contains reacts with oxygen (O2) to form carbon dioxide (CO2). (a) What is the theoretical yield of CO2 in grams from 0.50 mol of C? (b) What is the percent yield if 10.0 g of (CO2) are formed?
C(s) + O_{2}(g) \longrightarrow CO_{2}(g)
To answer part (a), convert the number of moles of reactant to the number of moles of product, as
in Section 5.8A. Then, convert the number of moles of product to the number of grams of product
using the product’s molar mass. This is the theoretical yield. To answer part (b), use the formula,
percent yield = (actual yield/theoretical yield) × 100%.
a. Calculate the theoretical yield using the procedure outlined in Sample Problem 5.16.
[1] Convert the number of moles of reactant to the number of moles of product using a mole–mole conversion factor.
• Use the coefficients in the balanced equation to write a mole–mole conversion factor for the two compounds—one mole of carbon forms one mole of CO2 . Multiply the number of moles
of reactant (carbon) by the conversion factor to give the number of moles of product (CO2) .
\begin{matrix} \text{Moles of reactant} & &\text{mole–mole conversion factor} & & \text{Moles of product} \\ \\ 0.50 \cancel{\text{ mol }}C & \times & \frac{1 \text{ mol }CO_{2}}{1 \cancel{\text{ mol }}C}&=& 0.50 \text{ mol } CO_{2} \end{matrix}
[2] Convert the number of moles of product to the number of grams of product—the theoretical yield—using the product’s molar mass.
• Use the molar mass of the product (CO_{2}, molar mass 44.01 g/mol) to write a conversion factor. Multiply the number of moles of product (from step [1]) by the conversion factor to give the number of grams of product.
\begin{matrix} \text{Moles of product} & &\text{molar mass conversion factor} & & \text{Grams of product} \\ \\ 0.50 \cancel{\text{ mol }}C & \times & \frac{44.01 \text{ g }CO_{2}}{1 \cancel{\text{ mol }}CO_{2}}&=& 22 \text{ g } CO_{2} \\ \\ &&&& \text{Theoretical yield} \\ &&&& \text{Answer: part (a)}\end{matrix}
b. Use the theoretical yield from part (a) and the given actual yield to calculate the percent yield.
\begin{matrix} \text{Percent yield} &= & \frac{\text{actual yield (g)}}{\text{theoretical yield (g)}}\times 100\% \\ \\ & =& \frac{10.0 \text{ g}}{22\text{ g}} \times 100\% & =& 45\% \\ \\ &&&& \text{Answer: part (b)}\end{matrix}