Question

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation: CH4(g)+H2O(g)→CO(g)+3H2(g) In a particular reaction, 25.0 L of methane gas (measured at a pressure of 732 torr and a temperature of 25 ∘C) is mixed with 22.2 L of water vapor (measured at a pressure of 704 torr and a temperature of 125 ∘C). The reaction produces 26.2 L of hydrogen gas measured at STPWhat is the percent yield of the reaction?

Answer 1

Answer:

The % yield of this reaction is 61.9 %

Explanation:

Step 1: Data given

Volume of methane = 25.0 L

Pressure of methane = 732 torr = 732 /760 atm = 0.9631579 atm

Temperature = 25.0 °C = 298 K

Volume of water vapor = 22.2 L

Pressure of water vapor = 704 torr = 704/760 atm = 0.92631579 atm

Temperature = 125 °C 398 K

The reaction produces 26.2 L hydrogen gas

Step 2: The balanced equation

CH4(g)+H2O(g)→CO(g)+3H2(g)

Step 3: Calculate moles methane

p*V = n*R*T

n  =(p*V)/(R*T)

⇒with n = the moles of methane = TO BE DETERMINED

⇒with p= the pressure of methane = 732 torr = 0.9631579 atm

⇒with V = the volume of methane = 25.0 L

⇒with R = the gas constant =0.08206 L*atm/mol*K

⇒with T = the temperature = 298 K

n = (0.9631579 * 25.0) / (0.08206*298)

n = 0.984668 moles

Step 4: Calculate moles H2O

p*V = n*R*T

n  =(p*V)/(R*T)

⇒with n = the moles of H2O= TO BE DETERMINED

⇒with p= the pressure of methane = 704 torr = 0.92631579  atm

⇒with V = the volume of methane = 22.2 L

⇒with R = the gas constant =0.08206 L*atm/mol*K

⇒with T = the temperature = 398 K

n = (0.92631579  * 22.2 )/(0.08206 * 398) = 0.62965 mol H2O

Step 5: Calculate moles H2

CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g)

For 1 mol CH4 we need 1 mol H2O to produce 1 mol CO and 3 moles H2

H2O is the limiting reactant. It will completely be consumed. (0.62965 moles). Methane is in excess. There will react 0.62965 moles. There will remain  0.984668 - 0.62965 = 0.355018 moles methane

For 0.62965 moles H2O we'll have 3*0.62965 = 1.88895 moles H2

Step 6: Calculate volume H2

p*V = n*R*T

V= (n*R*T)/p

⇒with V = the volume of H2 = TO BE DETERMINED

⇒with n = the moles of H2 produced = 1.88895 moles

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 273K

⇒with p = the pressure of H2 = 1.0 atm

V = (1.88895 * 0.08206 * 273) / 1.0

V = 42.32 L

Step 7: Calculate the percent yield

% yield = (actual yield / theoretical yield) * 100 %

% yield = (26.2 / 42.32) * 100 %

% yield = 61.9 %

The % yield of this reaction is 61.9 %