Given Problem: Methanol (CH3OH) can be synthesized as: CO+2H2→CH3OH. What volume in liters of H2(g) is needed to make 35.7 grams of CH3OH?
1. We are given the temperature at 355 K and the pressure is 738 mmHg.
2. First, you must convert the target mass of CH3OH to moles. To do this, you use the molar mass of CH3OH which is 32.04 grams/mol.
3. We can now find how much H2 gas is needed by using the molar ratio between it and CH3OH. In this case, for every two moles of hydrogen gas that react, one mole of CH3OH is produced. This can be interpreted as a 2:1 ratio- we use it to find that 2.228 moles of hydrogen gas are needed to react to produce 1.114 moles of CH3OH.
4. Convert your mmHg to atmospheres using 760 mmHg=1 atm.
5. The ideal gas law can be arranged to find volume. Plug in your moles, atmospheres, and the temperature in Kelvin. You can use either 8.314 J/Kg*mol or .08206 L*atm/mol*K for the ‘r’ constant. In this case, I use the .08206 value because it matches all my units.
6. The final answer is 66.853 L of H2. This is how many liters of hydrogen gas is needed to react with carbon monoxide to yield 35.7 grams of methanol.
Questions? Comments? Feedback? Leave a reply!