Question

Hydrogen gas can be formed by the reaction of methane with water according to the equation.

CH4 + H20 -> CO + 3H2

In a particular reaction, 25.5 L of methane gas( measure at a pressure of 732 torr and temperature of 25 C) mixes with 22.8 L of water vapor( measured at a pressure of 702 torr and a temperature of 125 C ) The reaction produces 26.2L of Hydrogen gas at STP. What is the percent yield of the reaction?

Answer 1

60.9 % is the yield of the reaction

Step-by-step 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%