Answer:
60.42% is the percent yield of the reaction.
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](https://tex.z-dn.net/?f=PV%3DnRT)
![n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol](https://tex.z-dn.net/?f=n%3D%5Cfrac%7BPV%7D%7BRT%7D%3D%5Cfrac%7B0.9542%20atm%5Ctimes%2026.0L%7D%7B0.0821%20atm%20L%2Fmol%20K%5Ctimes%20298.15%20K%7D%3D1.0135%20mol)
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'](https://tex.z-dn.net/?f=P%27V%27%3Dn%27RT%27)
![n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol](https://tex.z-dn.net/?f=n%27%3D%5Cfrac%7BPV%7D%7BRT%7D%3D%5Cfrac%7B0.9100%20atm%5Ctimes%2023.0L%7D%7B0.0821%20atm%20L%2Fmol%20K%5Ctimes%20398.15%20K%7D%3D0.6402%20mol)
![CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)](https://tex.z-dn.net/?f=CH_4%28g%29%2BH_2O%28g%29%5Crightarrow%20CO%28g%29%2B3H_2%28g%29)
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:
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:
![\frac{1}{22.4 L}\times 26 L= 1.1607 mol](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B22.4%20L%7D%5Ctimes%2026%20L%3D%201.1607%20mol)
Moles of hydrogen gas obtained experimentally = 1.1607 mol
Percentage yield of hydrogen gas of the reaction:
![\frac{Experimental}{Theoretical}\times 100](https://tex.z-dn.net/?f=%5Cfrac%7BExperimental%7D%7BTheoretical%7D%5Ctimes%20100)
![\%=\frac{ 1.1607 mol}{1.9208 mol}\times 100=60.42\%](https://tex.z-dn.net/?f=%5C%25%3D%5Cfrac%7B%201.1607%20mol%7D%7B1.9208%20mol%7D%5Ctimes%20100%3D60.42%5C%25)
60.42% is the percent yield of the reaction.