Question

If all the O-atoms from 4.4 g CO2, 6.022 × 1022 molecules of N2O5, 0.2 moles of CO and 1.12 L of
SO2 gas at NTP are removed and combined to form O2 gas, then the resulting gas occupies a
volume of ..... at NTP.
(A) 22.4 L (B) 44.8 L (C) 33.6 L (D) 11.2 L
plz answer it correctly

Answer 1

Given info : all the O-atoms from 4.4 g CO2, 6.022 × 1022 molecules of N2O5, 0.2 moles of CO and 1.12 L of SO2 gas at NTP are removed and combined to form O2 gas.

To find : the resulting gas occupies volume at NTP

solution : first you have to find total no of moles of atoms of oxygen. let's find it.

4.4 g of CO2 :

no of moles of CO2 = given mass/molecular mass

= 4.4/44 = 0.1 mol

we see, 2 atoms are present in 1 molecule of CO2 so, no of moles of O atoms in 0.1 mole of CO2 = 2 × 0.1 = 0.2 mol

6.022 × 10²² molecules of N2O5 :

no of moles of N2O5 = no of molecules/Avogadro's number

= 6.022 × 10²²/6.022 × 10²³ = 0.1

we see, 5 O atoms are present in 1 molecule of N2O5 so no of moles of O atoms in 0.1 mol of N2O5 = 5 × 0.1 = 0.5 mol

similarly , no of O atom in 0.2 mol of CO = 0.2

no of O atoms in 1.12L of SO2 = no of moles of SO2 × 2

= 1.12L/22.4L × 2

= 1/20 × 2

= 0.1

so total no of O atoms = 0.2 + 0.5 +0.2 + 0.1 = 1

we know, 2 moles of O atoms form 1 mole of O2 gas.

so, 1 mole of O atoms form 0.5 mol of O2 gas.

at NTP,

volume of 1 mole of gas = 22.4 L

so, volume of 0.5 mol of gas = 11.2 L

Therefore the volume of O2 gas at NTP is 11.2 L