What volume (in L) of water vapor will be produced from the reaction of 24.65 L of oxygen?
2C2H6(9) + 702(9) — 4CO2(g) + 6H20(9)
Enter

21.13 L

Explanation:

Step 1: Write the balanced equation

2 C₂H₆(g) + 7 O₂(g) ⇒ 4 CO₂(g) + 6 H₂O(g)

Step 2: Determine the appropriate volume ratio

Since all the gases are in the same container at the same temperature and pressure, the volume ratio is equal to the molar ratio, because the volume depends on the number of moles. The volume ratio of O₂(g) to H₂O(g) is 7:6.

Step 3: Determine the volume of H₂O produced from 24.65 L of O₂

24.65 L O₂ × 6 L H₂O/7 L O₂ = 21.13 L H₂O

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 :