Question 5.12: Methanol (CH3OH) can be synthesized by the reaction shown he...
Methanol (CH_{3}OH) can be synthesized by the reaction shown here:
CO(g) + 2 H_{2}(g) → CH_{3}OH(g)
What volume (in liters) of hydrogen gas, measured at a temperature of 355 K and a pressure of 738 mmHg, is required to synthesize 35.7 g of methanol?
| SORT You are given the mass of methanol, the product of a chemi-cal reaction. You are asked to find the required volume of one of the reactants (hydrogen gas) at a specified temperature and pressure. | GIVEN 35.7 g CH_{3}OH, T = 355 K, P = 738 mmHg FIND V_{H_{2}} |
| STRATEGIZE You can calculate the required volume of hydrogen gas from the number of moles of hydrogen gas, which you can obtain from the number of moles of methanol via the stoichiometry of the reaction. First, find the number of moles of methanol from its mass by using the molar mass. Then use the stoichiometric relationship from the balanced chem-ical equation to find the number of moles of hydrogen needed to form that quantity of methanol. Finally, substitute the number of moles of hydrogen as well as the pressure and temperature into the ideal gas law to find the vol-ume of hydrogen |
CONCEPTUAL PLAN g CH_{3}OH → mol CH_{3}OH . \frac{1 mol CH_{3}OH}{32.04 g CH_{3}OH} mol CH_{3}OH → mol H_{2} |
| SOLVE Follow the conceptual plan to solve the problem. Begin by using the mass of methanol to get the number of moles of methanol. Next, convert the number of moles of methanol to moles of hydrogen. Finally, use the ideal gas law to find the volume of hydrogen. Before substituting into the equation, convert the pressure to atmospheres. |
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35.7 \cancel{g CH_{3}OH} \times\frac{1 mol CH_{3}OH}{32.04 \cancel{g CH_{3}OH} }= 1.1\underline{1}42 mol CH_{3}OH
1.1\underline{1}42 \cancel{mol CH_{3}OH} \times\frac{2 mol H_{2}}{1 \cancel{mol CH_{3}OH}} = 2.2\underline{2}84 mol H_{2}
V_{H_{2}} = \frac{n_{H_{2}} RT}{P}
P = 738\cancel{ mmHg} \times\frac{1 atm}{760 \cancel{mmHg}} = 0.97\underline{1}05 atm
V_{H2} =\frac{(2.2\underline{2}84 \cancel{mol}) \left(0.08206\frac{L.\cancel{atm}}{\cancel{mol}.\cancel{K}} \right) (355 \cancel{K})}{0.97\underline{1}05 \cancel{atm}}
= 66.9 L
CHECK The units of the answer are correct. The magnitude of the answer (66.9 L) seems reasonable since you are given slightly more than 1 molar mass of methanol, which is therefore slightly more than 1 mol of methanol. From the equation you can see that 2 mol hydrogen is required to make 1 mol methanol. Therefore, the answer must be slightly greater than 2 mol hydrogen. Under standard temperature and pressure, slightly more than 2 mol hydrogen occupies slightly more than 2 × 22.4 L = 44.8 L. At a temperature greater than standard temperature, the volume is even greater; therefore, the magnitude of the answer is reasonable.