Question

copper (II) oxide reacts with hydrogen gas to produce copper and water, what volume of H2 at 1.01 at and 225C is needed to reduce 0.446 mol of copper (II) oxide ?

Answer 1

The balanced equation is : CuO + H₂ Cu + H₂O

From the balanced equation ,

1 mole of CuO is reduced by 1 mole of Hydrogen gas

0.446 mole of CuO is reduced by 0.446 mole of Hydrogen gas

Calculation of volume of Hydrogen gas :

We know that PV = nRT

Where

T = Temperature = 225 ^oC = 225+273 = 498 K

P = pressure = 1.01 atm atm

n = No . of moles = 0.446 mol

R = gas constant = 0.0821 L atm / mol - K

V = Volume of the H₂ gas = ?

Plug the values we get V = (nRT) / P

                                    = (0.446 x 0.0821x498) / 1.01

                                    = 18.05 L

Therefore the volume of H₂ gas needed is 18.05 L