Answer:
There will be produced 379 grams of H2
The percent yield under these conditions is 7.45 %
Explanation:
Step 1: The balanced equation
CH4(g)+CO2(g) → 2CO(g)+2H2(g)
Step 2: Given data
Kp = 4.5*10²
Temperature = 825 Kelvin
Volume = 85.0 L
Mass of CH4 = 22.3 Kg
Mass of CO2 = 55.4 Kg
Molar mass of CH4 = 16.04 g/mole
Molar mass of CO2 = 44.01 g/mole
Step 3: Calculate moles of CH4
moles of CH4 = mass of CH4 / Molar mass of CH4
moles of CH4 = 22300 / 16.04 = 1390.3
Step 4: Calculate moles of CO2
moles of CO2 = mass of CO2 / Molar mass of CO2
moles of CO2 = 55400 / 44.01 = 1258.80 moles
Step 5: Calculate moles of H2
For 1 mole of CH4 we need 1 mole of CO2 to produce 2 moles of H2
so there will be produced 2*1258.8 = 2517.6 moles of H2
Step 6: Calculate theoretical mass of H2
mass of H2 = Number of moles of H2 * Molar mass of H2
mass of H2 = 2517.6 moles *2.02 g/mole = 5085.552 grams
Step 7: Calculate pressure of CH4
P*V = n*R*T
P =(n*R*T) / V
P = 1390.3*.0821*825/85 = 1107.86atm
Step 8: Calculate pressure of CO2
P*V = n*R*T
P =(n*R*T) / V
P = 1258.80*.0821*825/85 = 1003.08atm
Step 9
Kp= 4.5 *100 =[P(H2)^2 * P(CO)^2]/[P(CO2) * P(CH4)]
450=16X^4/[(1107.86-X)(1003.08-X)]
450=16X^4/[(1107.86)(1003.08)]
(1107.86)(1003.08)*450=16X^4
X = 74.77
We plug this value in for "X" in the ICE chart. Only H2 since thats what the problem wants.
for H2 (2 moles) so 2X = 2*74.77 = 149.54
Step 10: Calculate number of moles of H2
n = P*V/ R*T
n = (149.54 *85 )/(0.0821 * 825) = 187.66 moles H2
Step 11: Calculate mass of H2
mass of H2 = Number of moles H2 * Molar mass of H2
mass of H2 = 187.66 * 2.02 g/moles = 379.07 grams H2 ≈ 379 grams of H2
Step 12: Calculate the yield
(379 grams of H2 / 5085.552 grams of H2 ) * 100 % = 7.45 %
The percent yield under these conditions is 7.45 %
