Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation: In a particular reaction, 25.5 L of methane gas (measured at a pressure of 732 torr and a temperature of 25 °C) is mixed with 22.0 L of water vapor (measured at a pressure of 704 torr and a temperature of 125 °C). The reaction produces 26.0 L of hydrogen gas measured at STP. What is the percent yield of the reaction?

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 %

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)

2b2t

explanation:

2b2t