Respuesta :
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
[tex]PV=nRT[/tex]
[tex]n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol[/tex]
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'
[tex]P'V'=n'RT'[/tex]
[tex]n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol[/tex]
[tex]CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)[/tex]
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:
[tex]\frac{3}{1}\times 0.6402 mol=1.9208 mol[/tex] 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:
[tex]\frac{1}{22.4 L}\times 26 L= 1.1607 mol[/tex]
Moles of hydrogen gas obtained experimentally = 1.1607 mol
Percentage yield of hydrogen gas of the reaction:
[tex]\frac{Experimental}{Theoretical}\times 100[/tex]
[tex]\%=\frac{ 1.1607 mol}{1.9208 mol}\times 100=60.42\%[/tex]
60.42% is the percent yield of the reaction.
