Question 7.15: On a space shuttle, LiOH is used to absorb exhaled CO2 from ...
On a space shuttle, LiOH is used to absorb exhaled CO_{2} from breathing air to form LiHCO_{3}.
LiOH(s) + CO_{2}(g) → LiHCO_{3}(s)What is the percent yield of LiHCO_{3} for the reaction if 50.0 g of LiOH gives 72.8 g of LiHCO_{3}?
STEP 1 State the given and needed quantities (moles).
| ANALYZE THE PROBLEM | Given | Need | Connect |
| 50.0 g of LiOH (reactant), 72.8 g of LiHCO_{3} (actual product) | percent yield of LiHCO_{3} | molar masses, mole-mole factor , percent yield expression | |
| Equation | |||
| LiOH(s) + CO_{2}(g) → LiHCO_{3}(s) | |||
STEP 2 Write a plan to calculate the theoretical yield and the percent yield.
Calculation of theoretical yield:
Calculation of percent yield:
Percent yield(%)= \frac{acntal yield of LiHCO_{3}}{theoretical yield of LiHCO_{3}} \times 100%
STEP 3 Use coefficients to write mole-mole factors; write molar mass factors.
\begin{array}{r c}\boxed{\begin{matrix} 1 mole of LiOH = 23.95 g of LiOH \\\frac{23.95 g LiOH}{1 mole LiOH} \text{ and } \frac{1 mole LiOH}{23.95 g LiOH} \end{matrix}} & \boxed{\begin{matrix} 1 mole of LiHCO_{3} = 67.96 g of LiHCO_{3} \\ \frac{67.96 g LiHCO_{3}}{ 1 mole LiHCO_{3}}\text{ and }\frac{1 mole LiHCO_{3}}{67.96 g LiHCO_{3}} \end{matrix}}\end{array}\begin{array}{r c}\boxed{\begin{matrix} 1 mole of LiHCO_{3}= 1 mole of LiOH \\\frac{1 mole LiHCO_{3}}{1 mole LiOH} \text{ and } \frac{1 mole LiOH}{1 mole of LiHCO_{3}} \end{matrix}}\end{array}
STEP 4 Calculate the percent yield by dividing the actual yield (given) by the theoretical yield and multiplying the result by 100%.
Calculation of theoretical yield:
Calculation of percent yield:
\frac{acntal yield (given) }{theoretical yield (calculated)} \times 100 % = \overset{Three Sfs}{\underset{Three Sfs}{\frac{72.8 \cancel{g LiHCO_{3}}}{142 \cancel{g LiHCO_{3}}}}} \times 100 % =\underset{Three Sfs }{51.3}%