Question

3. CH3OH is synthesized via the following reaction: CO (g) + 2H2 (g)  CH3OH (g) . If the reaction occurs at 748 mmHg and 25 °C , what volume of H2 gas (in L) is necessary to synthesize 25.8 g CH3OH? What volume (L) of CO gas is required?

4. Hydrogen gas can be formed by the reaction of methane and water according to the following equation: CH4 (g) + H2O (g)  CO (g) + 3H2 (g) . In one reaction, 25.5 L of methane gas (at P = 732 Torr, T = 25 °C) is mixed with 22.8 L of water vapor (at P = 702 Torr, T = 125 °C). The reaction produces 26.2 L hydrogen gas at STP. What is the percent yield of the reaction?

Answer 1

3.

CO + 2H2 -----> CH3OH

Mass of CH3OH = 25.8 g

moles of CH3OH = 25.8 / 32 = 0.806

1 mol of CH3OH requires 2 mol of H2

0.806 mol of CH3Oh requires ( 2* 0.806) = 1.612 mol

PV = nRT

P = 748 mmHg = 0.984 atm , T = 273 + 25 = 298 K

V = nRT/P = 1.612 * 0.0821 * 298 / 0.984 = 40.1 L

moles of CO = moles of CH3OH = 0.806 mol

V = nRT/P = 0.806 * 0.0821 * 298 / 0.984 = 20.04 L

4.

CH4 + H2O ------> CO + 3H2

For methane gas , P = 732 torr = 0.963 atm , T = 298 K

n = PV / RT = 0.963 * 25.5 / (0.0821 * 298) = 1 mol

For water vapour , P = 702 torr = 0.924 atm , T = 125 + 273 = 398 K

n = PV / RT = 0.924 * 22.8 / (0.0821*398) = 0.645 mol

Here 1 mol of H2O requires 1 mol of CH4 but moles of H2O available = 0.645 mol . So, H2O is limiting reagent.

1 mol of H2O gives 3 mol of H2

0.645 mol of H2O gives (3 * 0.645) = 1.934 mol

Volume of H2 produced at STP = 1.934 * 22.4 L = 43.33 L

Percentage Yield = Actual Yield * 100 / Theoretical yield

% yield = 26.2 * 100 / 43.33 = 60.47%