Answer:
60.9 % is the yield of the reaction
Explanation:
The reaction is: CH₄ + H₂O → CO + 3H₂
Let's apply the Ideal Gas Law to determine the moles of each reactant.
First of all, we convert the pressure from Torr to atm
732 Torr . 1atm / 760Torr = 0.963 atm Pressure of CH₄
702 Torr . 1atm / 760Torr = 0.923 atm Pressure of H₂O vapor
Now we can apply P . V = n . R . T
We convert T° from °C to K → 25°C + 273 = 298 K and 125°C + 273 = 398K
CH₄ → (25.5L . 0.963 atm) / 0.082 . 298K = n → 1 mol
H₂O vapor → (22.8L . 0.923 atm) / 0.082 . 398K = n → 0.64 mol
Ratio is 1:1, so the limiting reactant is the water vapor. I need 1 mol to react with 1 mol of methane but I only have 0.64 moles.
At the 100 % yield, 0.64 moles of water vapor may produce (0.64 . 3) = 1.92 moles of H₂.
We apply the P . V = n .R . T to determine the moles of H₂ produced at STP
1 atm . 26.2L = n . 0.082 . 273K
(1 atm . 26.2L) / (0.082 . 273K) = n → 1.17 moles of H₂
Percent yield of the reaction is (Yield produced / Theoretical yield) . 100
(1.17 moles / 1.92 moles) . 100 = 60.9%
