# 4H + 4e = 2H2 4OH = O2 + 2H2O +4e. The weight of the solution is 1400g. During electrolysis, 1.1682L gas was produced at 273.15K and 1atm. Calculate the weight (in g) of the oxygen gas evolved. Relative atomic weights: H:1 O:16.

0.27782g of oxygen is evolved.

The electrolysis of water is given by the equation:

`2 H_2O -> 2 H_2 + O_2`

When the given sample of water undergoes electrolysis, 1.1682 L of gas is produced at a temperature 273.15 K and pressure 1 atm.

The ideal gas equation gives PV = nRT, where P is the pressure in atm., V is the volume in liters, n is the number of moles of the gas, R is the ideal gas constant equal to 0.0821 atm•L/mol•K T, and T is the temperature in kelvin.

If 1.1682 L is the volume of n moles of gas, n is equal to

n = (PV)/(RT) = (1*1.1682)/(0.0821*273.15)

n = 0.05209

The electrolysis of n moles of water leads to the formation of n moles of hydrogen gas and n/2 moles of oxygen gas. Totally, 1.5*n moles of gas is produced.

As n = 0.05209, the moles of oxygen evolved is 0.017364.

The molar mass of oxygen is 16.

0.017364 moles of oxygen is equivalent to 0.27782 g of oxygen.

The mass of oxygen evolved in grams is 0.27782.

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