Answer:
The percent yield of the reaction is 62.05 %
Explanation:
Step 1: Data given
Volume of methane = 25.5 L
Pressure of methane = 732 torr
Temperature = 25.0 °C = 298 K
Volume of water vapor = 22.0 L
Pressure of H2O = 704 torr
Temperature = 125 °C
The reaction produces 26.0 L of hydrogen gas measured at STP
Step 2: The balanced equation
CH4(g) + H2O(g) → CO(g) + 3H2(g)
Step 3: Calculate moles methane
p*V = n*R*T
⇒with p = the pressure of methane = 0.963158 atm
⇒with V = the volume of methane = 25.5 L
⇒with n = the moles of methane = TO BE DETERMINED
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 298 K
n = (p*V) / (R*T)
n = (0.963158 * 25.5 ) / ( 0.08206 * 298)
n = 1.0044 moles
Step 4: Calculate moles H2O
p*V = n*R*T
⇒with p = the pressure of methane = 0.926316 atm
⇒with V = the volume of methane = 22.0 L
⇒with n = the moles of methane = TO BE DETERMINED
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 398 K
n = (p*V) / (R*T)
n = (0.926316 * 22.0) / (0.08206 * 398)
n = 0.624 moles
Step 5: Calculate the limiting reactant
For 1 mol methane we need 1 mol H2O to produce 1 mol CO and 3 moles H2
H2O is the limiting reactant. It will completely be consumed (0.624 moles).
Methane is in excess. There will react 0.624 moles. There will remain 1.0044 - 0.624 moles = 0.3804 moles methane
Step 6: Calculate moles hydrogen gas
For 1 mol methane we need 1 mol H2O to produce 1 mol CO and 3 moles H2
For 0.624 moles H2O we'll have 3*0.624 = 1.872 moles
Step 9: Calculate volume of H2 at STP
1.0 mol at STP has a volume of 22.4 L
1.872 moles has a volume of 1.872 * 22.4 = 41.9 L
Step 10: Calculate the percent yield of the reaction
% yield = (actual yield / theoretical yield) * 100 %
% yield = ( 26.0 L / 41.9 L) *100 %
% yield = 62.05 %
The percent yield of the reaction is 62.05 %
