(3)
how many liters are in 1.35 x 105 grams of oxygen at stp?

V = 8.97×10⁴ L

Explanation:

Given data:

Mass of oxygen = 1.35×10⁵ g

Litters of oxygen = ?

Solution:

First of all we will calculate the number of moles.

Number of moles = mass/ molar mass

Number of moles = = 1.35×10⁵ g / 32 g/mol

Number of moles = 0.04 ×10⁵ mol

Litters of oxygen:

PV = nRT

V = nRT/P

V = 0.04 ×10⁵ mol × 0.0821 atm. L/ mol. K ×273 K / 1atm

V = 0.897×10⁵ atm.L / 1atm

V = 8.97×10⁴ L

9.46 x 10⁴ L

Explanation:

Data Given:

mass of oxygen (O₂) = 1.35 x 10⁵

Liters of  oxygen O₂ (volume) = ?

Solution:

First find no. of moles of oxygen

Formula used

            no. of moles = mass in gram / molar mass

molar mass of oxygen (O₂) = 2(16)

molar mass of oxygen (O₂) = 32 g/mol

Put values in above formula

             no. of moles = 1.35 x 10⁵ / 32 g/mol

             no. of moles = 4219 mol

Now we will find volume of oxygen

Formula used

            no. of moles = Volume of gas / Molar volume

Rearrange the above equation to find volume

        Volume of gas = no. of moles x Molar Volume . . . . . . (1)

Where

Molar volume = 22.42 L /mol

Put values in above equation 1

             Volume of gas = 4219 mol x 22.42 L /mol

             Volume of gas = 94584 L

             Volume of gas = 9.46 x 10⁴ L

So,

Volume of oxygen (O₂) = 9.46 x 10⁴ L

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 :