Question

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:
`CH_4(g) + H_2O(g) \rightarrow CO(g) + 3H_2(g)`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?

Answer 1

62.02 %

Step-by-step explanation:

The percent yield of the reaction can be calculated using the following equation:

`\% = (y_(e))/(y_(t))*100`

Where:

`y_(e)`: is the experimental yield

`y_(t)`: is the theoretical yield

We can see that we need to find the theoretical yield and the experimental yield.

To calculate the theoretical yield we need to find the number of moles of the reactants using the Ideal Gas Law:

`n_{CH_(4)} = (PV)/(RT)`

Where:

P: is the pressure of the gas = 732 torr

V: is the volume of the gas = 25.5 L

R: is the gas constant = 0.082 L*atm/(K*mol)

T: is the temperature = 25 °C

Hence the number of moles of the methane is:

`n_{CH_(4)} = (732 torr \cdot (1 atm)/(760 torr)*25.5 L)/(0.082 L*atm*K^(-1)*mol^(-1)*(25 + 273 K)) = 1.005 moles`

Similarly, the number of moles of the water vapor is:

`n_{H_(2)O} = (704 torr \cdot (1 atm)/(760 torr)*22.0 L)/(0.082 L*atm*K^(-1)*mol^(-1)*(125 + 273 K)) = 0.624 moles`

Now, we need to find the limiting reactan. In the following equation:

CH₄(g) + H₂O(g) → CO(g) + 3H₂(g)

we have that 1 mol of CH₄ reacts with 1 mol of H₂O:

`(1 mol CH_(4))/(1 mol H_(2)O)*0.624 moles H_(2)O = 0.624 moles CH_(4)`

We need 0.624 moles of CH₄ to react with H₂O, and we have 1.005 moles of CH₄. Therefore, the limiting reactant is the H₂O.

Since 1 mol of H₂O produces 3 moles of H₂, the number of H₂ moles produced is:

`(3 mol H_(2))/( 1 mol H_(2)O)*0.624 moles H_(2)O = 1.872 moles`

The experimental yield is:

`n = (PV)/(RT) = (1 atm*26.0 L)/(0.082 L*atm*K^(-1)*mol^(-1)*273 K) = 1.161 moles`

Finally, the percent yield of the reaction is:

`% y = (26.0 L)/(41.91 L)* 100 = 62.04 /%`
`\% y = (1.161)/(1.872)*100 = 62.02 \%`

I hope it helps you!

Answer 2

The percent yield of the reaction is 62.05 %

Step-by-step 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 %