To synthesize 0.55 moles of [tex]\( CH_3OH \)[/tex] through the given reaction:
[tex]\[ CO(g) + 2H_2(g) \longrightarrow CH_3OH(g) \][/tex]
we need to follow a step-by-step process to determine the volumes of [tex]\( H_2 \)[/tex] and [tex]\( CO \)[/tex] gases required, considering the provided conditions of 748 mm Hg pressure and [tex]\( 86^\circ C \)[/tex] temperature.
### Step 1: Determine the moles of hydrogen ([tex]\( H_2 \)[/tex])
According to the balanced chemical equation, 2 moles of [tex]\( H_2 \)[/tex] are required to produce 1 mole of [tex]\( CH_3OH \)[/tex]. Therefore, to produce 0.55 moles of [tex]\( CH_3OH \)[/tex]:
[tex]\[ \text{Moles of } H_2 = 2 \times 0.55 = 1.1 \text{ moles of } H_2 \][/tex]
### Step 2: Determine the volume of hydrogen ([tex]\( H_2 \)[/tex])
Using the Ideal Gas Law:
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure (in atm),
- [tex]\( V \)[/tex] is the volume (in liters),
- [tex]\( n \)[/tex] is the number of moles,
- [tex]\( R \)[/tex] is the Ideal Gas Constant ([tex]\( 0.0821 \, \text{L·atm/(mol·K)} \)[/tex]),
- [tex]\( T \)[/tex] is the temperature (in Kelvin).
#### Convert the pressure from mm Hg to atm:
[tex]\[ P = \frac{748 \, \text{mm Hg}}{760} \approx 0.984 \, \text{atm} \][/tex]
#### Convert the temperature from Celsius to Kelvin:
[tex]\[ T = 86 + 273.15 = 359.15 \, \text{K} \][/tex]
Now, substitute the values into the Ideal Gas Law to find the volume ([tex]\( V \)[/tex]) for [tex]\( H_2 \)[/tex]:
[tex]\[ V_{H_2} = \frac{(1.1 \, \text{moles}) \times (0.0821 \, \text{L·atm/(mol·K)}) \times (359.15 \, \text{K})}{0.984 \, \text{atm}} \][/tex]
[tex]\[ V_{H_2} \approx 32.96 \, \text{liters} \][/tex]
### Step 3: Determine the moles of carbon monoxide ([tex]\( CO \)[/tex])
According to the balanced chemical equation, 1 mole of [tex]\( CO \)[/tex] is required to produce 1 mole of [tex]\( CH_3OH \)[/tex]. Therefore, to produce 0.55 moles of [tex]\( CH_3OH \)[/tex]:
[tex]\[ \text{Moles of } CO = 0.55 \text{ moles of } CO \][/tex]
### Step 4: Determine the volume of carbon monoxide ([tex]\( CO \)[/tex])
Using the Ideal Gas Law:
[tex]\[ V_{CO} = \frac{(0.55 \, \text{moles}) \times (0.0821 \, \text{L·atm/(mol·K)}) \times (359.15 \, \text{K})}{0.984 \, \text{atm}} \][/tex]
[tex]\[ V_{CO} \approx 16.48 \, \text{liters} \][/tex]
### Summary
- The amount of [tex]\( H_2 \)[/tex] required is 1.1 moles, which is approximately 32.96 liters under the given conditions.
- The amount of [tex]\( CO \)[/tex] required is 0.55 moles, which is approximately 16.48 liters under the same conditions.
