Question

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)+3H2(g)CH4(g)+H2O(g)→CO(g)+3H2(g) In a particular reaction, 26.5 LL of methane gas (measured at a pressure of 736 torr and a temperature of 25 ∘C) is mixed with 22.6 L of water vapor (measured at a pressure of 702 torr and a temperature of 125 ∘C). The reaction produces 26.2 L of hydrogen gas at STP. Part A What is the percent yield of the reaction?

Answer 1

  CH₄(g)+H₂O(g)→CO(g)+3H₂(g)

According to ideal gas equation :

methane : PV = nRT

P = 736 torr = 0.968 atm (760 torr = 1 atm)

T = 25 C = 25+273 = 298 K

0.968 atm x 26.5 l = n X 0.0821 lt atm/mol K x 298 K

25.652/24.47 = n

n = 1.048 moles

water : PV = nRT

P = 702 torr = 0.924 atm (760 torr = 1 atm)

T = 125 C = 125+273 = 398 K

0.924 atm x 22.6 l = n X 0.0821 lt atm/mol K x 398 K

20.88/32.68 = n

n = 0.639 moles

Molar ratio of methane and water vapor = 1:1

1 mole of methane require 1 mole of water vapor

1.048 moles of methane require 1.048 moles of water vapor but 0.639 moles of water vapor is present

So limiting reactant is water vapor

given mass of water vapor = moles x molar mass = 0.639 mol x 18.02 g/mol = 11.5 g

Molar ratio of water vapor and hydrogen gas = 1:3

So theoretical moles of hydrogen = 3 x 0.639 mol = 1.917 moles

Theoretical mass of hydrogen = moles x molar mass = 1.917 mol x 2.016 g/mol = 3.865 gms

practical moles of hydrogen :

PV = nRT

1 atm x 26.2 l = n x 0.0821 lt atm/mol K x 273 K

26.2/22.4 = n

n = 1.1696 moles

Practial mass of hydrogen gas = moles x molar mass = 1.1696 mol x 2.016 g/mol = 2.358 g

Percent yield = (practical yield/theoretical yield) x 100 = 2.358/3.865) x 100 = 63.99%