The volumetric flow rate of aniline is 1500 cm/min. what is the mass flow rate in g/sec? what is the molar flow rate in kmol/min? (2) the volumetric flow rate of a 4m ammonium nitrate aqueous solution is 450 m/hr. the specific gravity of the solution is 1.54 what is the mass fraction of ammonium nitrate?

1) a) mass flow = 25.543 g/s

b) molar flow = 0.0165 Kmol/min

2) X NH4NO3 = 0.208

Explanation:

∴ ρaniline = 1.0217 g/cm³

⇒ mass flow = 1500 cm³/min * 1.0217 g/cm³ * min/60s

⇒ mass flow = 25.543 g/s

b) molar flow:

∴ Mw aniline = 93.13 g/mol

⇒ molar flow = (25.543 g/s) * (mol/ 93.13g) * (Kmol / 1000 mol) * (60 s/min)

⇒ molar flow = 0.0165 Kmol/min

2) vol. flow = 450 m³/h

∴ ρr = 1.54 = ρNH4NO3 / ρH2O

∴ ρH2O = 1 g/mL = 1000 Kg/m³

⇒ ρNH4NO3 = 1.54 * ( 1000 Kg/m³ ) = 1540 Kg/m³

⇒ mass flow NH4NO3 = 1540 Kg/m³ * 450 m³/h = 693000 Kg/h

∴ C NH4NO3 = 4 M = 4 mol/L * ( 1000 L/m³ ) = 4000 mol/m³

∴ Mw NH4NO3 = 80.043 g/mol

⇒ volumetric mass NH4NO3 (aq) = 4000 mol/m³ * 80.043 g/mol = 320172 g/m³ = 320.17 Kg/m³

∴ mass flow NH4NO3 (aq) = 320.17 Kg/m³ * 450 m³/h = 144077.4 Kg/h

⇒ mass fraction (X) =  mass flow NH4NO3(aq) / mass flow NH4NO3

⇒ X = 144077.4 / 693000

⇒ X = 0.208

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

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explanation: