Question

CH3OH can be synthesized by the reaction:

CO(g)+2H2(g)→CH3OH(g) How many liters of H2 gas, measured at 746 mmHg and 86 ∘C, are required to synthesize 0.61 mol of CH3OH?

Express your answer using two significant figures.

How many liters of CO gas, measured under the same conditions, are required?
Express your answer using two significant figures.

Answer 1

Answer: To determine the number of liters of H2 gas required to synthesize 0.61 mol of CH3OH, we'll use the ideal gas law equation:

PV = nRT

Where:

P = Pressure in atm (converted from 746 mmHg)

V = Volume in liters (unknown)

n = Number of moles of gas (0.61 mol)

R = Ideal gas constant (0.0821 L·atm/(mol·K))

T = Temperature in Kelvin (converted from 86 °C)

First, let's convert the temperature from Celsius to Kelvin:

T(K) = T(°C) + 273.15

T(K) = 86 + 273.15

T(K) = 359.15 K

Now, we can rearrange the ideal gas law equation to solve for V:

V = (nRT) / P

Plugging in the values:

V = (0.61 mol * 0.0821 L·atm/(mol·K) * 359.15 K) / (746 mmHg * 1 atm/760 mmHg)

Simplifying:

V ≈ 0.148 L

Therefore, approximately 0.148 liters of H2 gas are required to synthesize 0.61 mol of CH3OH.

To find the number of liters of CO gas required under the same conditions, we use the stoichiometric coefficients of the balanced equation:

1 mol CO : 2 mol H2

Since we know the number of moles of CH3OH (0.61 mol), we can deduce that half that amount of moles of CO gas is required:

0.61 mol CH3OH * (1 mol CO / 2 mol CH3OH) = 0.305 mol CO

Using the same ideal gas law equation, we can find the volume of CO gas:

V = (nRT) / P

V = (0.305 mol * 0.0821 L·atm/(mol·K) * 359.15 K) / (746 mmHg * 1 atm/760 mmHg)

Simplifying:

V ≈ 0.074 L

Therefore, approximately 0.074 liters of CO gas are required under the same conditions.