12.8 L of oxygen gas contains 0.0498 moles. At constant temperature and pressure, what volume does 0.0136 moles of the gas fill?

V₂ = 3.49 L

Explanation:

Given data:

Initial volume of oxygen = 12.8 L

Number of moles = 0.0498 mol

Final volume = ?

Final number of moles = 0.0136 mol

Solution:

According to Avogadro law the number of moles of gas and volume are directly related to each other at constant temperature and pressure.

V₁/n₁  = V₂/n₂

Now we will put the values in formula.

12.8 L /0.0498 mol = V₂/ 0.0136 mol

V₂ = 12.8 L× 0.0136 mol/0.0498 mol

V₂ = 0.174 L.mol/0.0498 mol

V₂ = 3.49 L

i think so it’s will be b yes i believe its

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 :