Question

A 0.50 molar aqueous solution of hydrochloric acid (HCl) flows into a process unit at 25°C and a rate of 1.25 m3/min. If the specific gravity of the HCl solution is 1.03 at 25°C determine: a) HCl concentration in kg/m3 b) Mass flow rate in kg HCl/s c) MassfractioninkgHCl/kgsolution d) Mole ratio in mol HCl/mol water e) Molar flow rate in kmol HCl/s

Answer 1

a) 18,23 kgHCl/m³

b) 0,38 kg HCl/s

c) 0,018 kgHCl/kgsolution

d) 8,75x10⁻³ molHCl/molH₂O

e) 0,010 kmolHCl/s

Step-by-step explanation:

The HCl solution is 0,50 mol/L; it flows in a rate of 1,25 m³/min; Its density is 1,03 kg/L. HCl weights 36,46 g/mol

a) The HCl concentration in kg/m³ is:

`(0,50molHCl)/(L)`×
`(36,46gHCl)/(mol)`×
`(1kg)/(1000g)`×
`(1000L)/(1m^3)`= 18,23 kg HCl/m³

b) As you have a concentration of 18,23 kgHCl/m³

`(18,23 kgHCl)/(m^3)`×
`(1,25m^3)/(min)`×
`(1min)/(60s)` = 0,38 kg HCl/s

c) The mass fraction in kgHCl/kgsolution is:

`(18,23kgHCl)/(m^3)`×
`(1m^3)/(1000L)`×
`(1L)/(1,03kg)` = 0,018 kgHCl/kgsolution

d) Supposing L of solution are just of water:

`(0,50 moleHCl)/(Lsolution)`×
`(1L)/(1,03kg)`×
`(1kg)/(1000g)`×
`(18,02gH_(2)O)/(1mole)`= 8,75x10⁻³ molHCl/molH₂O

e) As the flow is 0,38kg HCl/s:

`(0,38 kg)/(s)`×
`(1kmole)/(36,46kgHCl)`= 0,010 kmolHCl/s

I hope it helps!