79.2 L of a gas are contained at STP. How many moles are present in the container?

3.54 mol

Explanation:

When at STP (Pressure of 1 atm & T of 273.16 K), one mol of any given gas occupies 22.4 liters.

With the above information in mind we can calculate the number of moles, using the factor 22.4 L/mol:

79.2 L ÷ 22.4 L/mol = 3.54 mol

There are 3.54 moles of the gas present in the container.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :