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 %
