Question

2H2O(l)→2H2(g)+O2(g)

What mass of H2O is required to form 1.4 L of O2 at a temperature of 335 K and a pressure of 0.916 atm ?

Answer 1

use ideal gas equation to calculate mole of O₂

Use ideal gas equation for calculation of mole of gas

Ideal gas equation

PV = nRT             where, P = atm pressure = 0.916 atm,

V = volume in Liter = 1.4 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 335 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.916X 1.4) / (0.08205 X 335) = 0.0467 mole

0.0467 mole of O₂

According to reaction 2 mole of H₂O produce 1 mole of O₂ then to produce 0.0467 mole of O₂ required H₂O =

0.0467 X 2 = 0.0933 mole of H₂O

molar mass of H₂O = 18.01528 gm / mol then 0.0933 mole of H₂O = 0.0933 X 18.01528 = 1.68 gm

1.68 gm of H₂O is required to form 1.4 L of O2 at a temperature of 335 K and a pressure of 0.916 atm