1000 ml of oxygen at ntp passed through siemens ozoniser when the resulting volume was 888 ml at ntp .quantity of ozonised oxygen is passed through excess of potassium iodide solution calculate weight of iodine liberated?

To calculate the weight of iodine liberated, we need to follow the stoichiometry of the reaction between ozone and potassium iodide.

The balanced chemical equation for the reaction is:

2 O3 + 2 KI → 2 KIO3 + I2

From the equation, we can see that for every 2 moles of ozone (O3) consumed, 1 mole of iodine (I2) is liberated.

First, we need to determine the number of moles of ozone that reacted. We can use the ideal gas law to do this.

Given:
Initial volume of oxygen = 1000 mL
Final volume of ozonized oxygen = 888 mL

Since both volumes are given at NTP (Normal Temperature and Pressure), we can assume the ideal gas behavior and use the relationship:

n = PV/RT

Where:
P = Pressure (in atm) = 1 atm (at NTP)
V = Volume (in liters)
R = Ideal gas constant = 0.0821 L·atm/(mol·K)
T = Temperature (in Kelvin) = 273 K (at NTP)

Initial number of moles of oxygen:
n1 = (1 atm) * (1000 mL / 1000) / (0.0821 L·atm/(mol·K) * 273 K)
= 0.048 moles

Final number of moles of ozonized oxygen:
n2 = (1 atm) * (888 mL / 1000) / (0.0821 L·atm/(mol·K) * 273 K)
= 0.042 moles

Since the ozonizer reacts excess oxygen to produce ozone, we can assume that the difference in moles, n1 - n2, indicates the moles of ozone reacted:

Moles of ozone reacted:
n_ozone = n1 - n2
= 0.048 moles - 0.042 moles
= 0.006 moles

Now that we know the number of moles of ozone, we can calculate the weight of iodine liberated using the stoichiometric ratio from the balanced equation.

Molar mass of iodine (I2) = 2 * Atomic mass of I (iodine) = 2 * 126.9 g/mol = 253.8 g/mol

Weight of iodine liberated:
Weight_I2 = n_ozone * (2 mol I2 / 2 mol O3) * (253.8 g/mol)
= 0.006 moles * 253.8 g/mol
= 1.52 g

Therefore, approximately 1.52 grams of iodine would be liberated in the given reaction.