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, 26.0 L of methane gas (measured at a pressure of 734 torr and a temperature of 25 ∘C) is mixed with 23.0 L of water vapor (measured at a pressure of 700 torr and a temperature of 125 ∘C). The reaction produces 26.0 L of hydrogen gas measured at STP. Part A What is the percent yield of the reaction? %

Answer 1

60.42% is the percent yield of the reaction.

Step-by-step explanation:

Moles of methane gas at 734 Torr and a temperature of 25 °C.

Volume of methane gas = V = 26.0 L

Pressure of the methane gas = P = 734 Torr = 0.9542 atm

Temperature of the methane gas = T = 25 °C = 298.15 K

Moles of methane gas = n

`PV=nRT`

`n=(PV)/(RT)=(0.9542 atm* 26.0L)/(0.0821 atm L/mol K* 298.15 K)=1.0135 mol`

Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

`P'V'=n'RT'`

`n'=(PV)/(RT)=(0.9100 atm* 23.0L)/(0.0821 atm L/mol K* 398.15 K)=0.6402 mol`

`CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)`

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

`(3)/(1)* 0.6402 mol=1.9208 mol` of hydrogen gas

Moles of hydrogen gas obtained theoretically = 1.9208 mol

The reaction produces 26.0 L of hydrogen gas measured at STP.

At STP, 1 mole of gas occupies 22.4 L of volume.

Then 26 L of volume of gas will be occupied by:

`(1)/(22.4 L)* 26 L= 1.1607 mol`

Moles of hydrogen gas obtained experimentally = 1.1607 mol

Percentage yield of hydrogen gas of the reaction:

`(Experimental)/(Theoretical)* 100`

`\%=( 1.1607 mol)/(1.9208 mol)* 100=60.42\%`

60.42% is the percent yield of the reaction.