Calculate the mass in grams of 8.24 x 1024 molecules of CO2

602.25 gm

Explanation:

1 mole of any gas contains 6.02 x 10²³ molecules of gas

6.02 x 10²³ molecules of CO₂ contains 1 mole of CO₂

8.24 x 10²⁴ molecules of CO₂ will contain 8.24 x 10²⁴/ 6.02 x 10²³ moles at STP

13.6877 moles

mol weight of CO₂ = 44 g

13.6877 moles of  CO₂ = 44 g  x 13.6877 g =  602.25 gm.

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 :

6.4 cm (max) - 25 cm (min)