Problem: Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:CH4(g) + H2O(g) → CO(g) + 3 H2 (g)In a particular reaction, 26.0 L of methane gas (measured at a pressure of 734 torr and a temperature of 25 oC) is mixed with 22.6 L of water vapor (measured at a pressure of 704 torr and a temperature of 125 oC). The reaction produces 26.6 L of hydrogen gas measured at STP.What is the percent yield of the reaction?

FREE Expert Solution

CH4(g) + H2O(g) → CO(g) + 3 H2 (g)

Calculate how much H2(g) is produced from each reactant:

PV=nRTn=PVRT


H2(g) from CH4(g):

P=734 torr×1 mmHg760 torr

P = 0.9658 atm

V = 26.0 L
T = 25°C + 273.15 = 298.15 K

nCH4=(0.9658 atm)(26.0 L)(0.08206L·atmmol·K)(298.15 K)

nCH4 = 1.0263 mol

moles H2=1.0263 mol CH4×3 mol H21 mol CH4

moles H2 = 3.079 mol


H2(g) from H2O(g):

22.6 L of water vapor (measured at a pressure of 704 torr and a temperature of 125 oC

P=704 torr×1 mmHg760 torr

P = 0.9263 atm

V = 22.6 L
T = 125°C + 273.15 = 398.15 K

nH2O=(0.9263 atm)(22.6 L)(0.08206L·atmmol·K)(398.15 K)

nH2O = 0.6407 mol

moles H2=0.6407 mol H2O×3 mol H21 mol H2O

moles H2 = 1.922 mol


H2O(g): produces a smaller amount of H2(g) → H2O(g) is the limiting reactant

theoretical yield = 1.922 mol H2


Calculate the theoretical yield of H2(g) (volume):

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Problem Details

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:
CH4(g) + H2O(g) → CO(g) + 3 H2 (g)
In a particular reaction, 26.0 L of methane gas (measured at a pressure of 734 torr and a temperature of 25 oC) is mixed with 22.6 L of water vapor (measured at a pressure of 704 torr and a temperature of 125 oC). The reaction produces 26.6 L of hydrogen gas measured at STP.

What is the percent yield of the reaction?

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