Question

Find the mass of each element in 7.84g of FeSO4.(NH4)2SO4.6H2O. What will be the volume of O2 at NTP in this sample

Answer 1

Molar mass of compound ​FeSO4(NH4)2SO4.6H2O = 392.14 g mol-1
Number of moles of compound in 7.84g sample = 7.84 g/ 392.14 g mol-1
= 0.02 mole
1 mole of sample contain 1 mole of Fe
So, 0.02 mole of sample would contain 0.02 moles of Fe.
Mass of Fe equivalent to 0.02 mole = 0.02 mol x molar mass of Fe
= 0.02 mol x 55.84 g mol-1
= 1.12 g

1 mole of sample contain 20 moles of H Atoms
So, 0.02 mole of sample would contain 0.02 x 20 = 0.4 moles of H​
Mass of H equivalent to 0.4 mole = 0.4 mol x molar mass of H
= 0.4 mol x 1 g mol-1
= 0.4 g​

1 mole of sample contain 2 moles of N Atoms
So, 0.02 mole of sample would contain 0.02 x 2 = 0.04 moles of N
Mass of N equivalent to 0.04 mole = 0.04 mol x molar mass of N
= 0.04 mol x 14 g mol-1
= 0.56 g​​
1 mole of sample contain 14 moles of O Atoms
So, 0.02 mole of sample would contain 0.02 x 14 = 0.28 moles of O
Mass of O equivalent to 0.28 mole = 0.28 mol x molar mass of O
= 0.28 mol x 16 g mol-1
= 4.48 g​​​
1 mole of sample contain 2 moles of S Atoms
So, 0.02 mole of sample would contain 0.02 x 2 = 0.04 moles of S
Mass of S equivalent to 0.04 mole = 0.04 mol x molar mass of S
= 0.04 mol x 32 g mol-1
= 1.28 g​​​

NTP is defined as normal temperature and pressure. For NTP conditions, the temperature is 293.15 K (20 0 C) and pressure is 1 atm pressure.
Putting the values of P, T ,R and n in formula PV = nRT we will get the volume of O2 in sample
1 Mole of O2 contains 2 moles of O atoms
So 0.28 Mole O atom will form = 0.14 mole O2
V = nRT/P
= 0.14 mol x 0.0821 L atm mol-1 K-1 x 293. 15 K
= 3.36 L