Question

A mixture of oxygen and hydrogen is analyzed by passing it over hot copper oxide and through a drying tube. Hydrogen reduces the CuO to metallic Cu. Oxygen then reoxidizes the copper back to CuO. We know that 100 cm3 of the mixture measured at 25o C and 750 mm yields 84.5 cm3 of dry oxygen measured at at 25o C and 750mm after passage over CuO and the drying agent. What is the original composition of the mixture? (Hint: First write balanced chemical equation for the reactions.)

Answer 1

0,42x10⁻³ moles of H₂ and 3,62x10⁻³ moles of O₂

Step-by-step explanation:

The reactions are:

H₂ + CuO → H₂O + Cu

2 Cu + O₂ → 2 CuO

The sum of these reactions is:

2 H₂ + O₂ → 2 H₂O

Total initial moles n are obtained with ideal gas law, thus:

`(P.V)/(R.T) = n`

Where

P is pressure: 750mm Hg ≡ 0,987 atm

V is volume: 0,100 L

R is gas constant: 0,082
`(atm.L)/(mol.K)`

T is temperature: 25°C ≡ 298,15K

Thus, n = 4,04x10⁻³ initial total moles

As netiher temperature or pressure changes at the end of the reaction it is possible to obtain O₂ moles thus:

4,04x10⁻³ moles ×
`(84,5 mL)/(100 mL)` = 3,41x10⁻³ moles O₂

Thus, H₂O moles are: 0,63x10⁻³ moles

As the reaction consumes 2 moles of H₂ and 1 mole of O₂ to produce 2 moles of H₂O. The moles of H₂O are ²/₃ of H₂ and ¹/₃ of O₂

The initial moles of H₂ are: 0,63x10⁻³ × ²/₃ = 0,42x10⁻³ moles of H₂

The initial moles of O₂ are: 3,41x10⁻³ + 0,63x10⁻³ × ¹/₃ = 3,62x10⁻³ moles of O₂

I hope it helps!