Question

Find the mass of each element in 7.84 g offeso4.(nh4)2so4.6h2o

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

Answer 2

The answer is provided in the attachment.

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