Question

Consider the following reaction: 2NiO(s)→2Ni(s)+O2(g) Vapor Pressure of Water versus Temperature Temperature (∘C) Pressure (mmHg) Temperature (∘C) Pressure (mmHg) 0 4.58 55 118.2 5 6.54 60 149.6 10 9.21 65 187.5 15 12.79 70 233.7 20 17.55 75 289.1 25 23.78 80 355.1 30 31.86 85 433.6 35 42.23 90 525.8 40 55.40 95 633.9 45 71.97 100 760.0 50 92.6

Part A If O2 is collected over water at 40.0 ∘C and a total pressure of 742 mmHg , what volume of gas will be collected for the complete reaction of 25.09 g of NiO? V V = L

Answer 1

Balanced equation:
2 NiO(s) ====> 2 Ni(s) + O₂(g)
Mass of NiO = 25.09 gm

Molar mass of NiO = 74.69 g/mol

Moles of NiO = 25.09 g / 74.69 g /mol = 0.3359 Moles

Moles of O2 produced =  0.1679 Moles

Using the mole and following formula we can calculate the volume of oxygen.

PV= nRT

P = Pressure in atm                       V= Volume in Liter

n = no of moles               R = 0.08206 L atm K-1 Mol-1

T = Temperature in Kelvin

P = 742 - 55.4 = 686.6 mmHg = 0.90342 atm V = ? T = 273 + 40 = 313 K n = 0.1679 Moles

V = nRT / P = 0.1679 Moles x 0.08206 L atm K-1 Mol-1 x 313 K / 0.90342 atm

V = 4.7734 Liter

Hence 4.7734 Liter of O2 gas will be produced.