Question

Se trateaza 2,35g de oxid de potasiu cu 50g de solutie HCl 8% Calculeaza:
Masa de sare obtinuta
Compozitia procentuala de masa a solutiei finale
Ajutati ma va rog

Answer 1

MK2O = 94g/mol

n=m/M=2,35/94 = 0,025 moli K2O

C=md/ms*100 => md=C*ms/100 = 8*50/100 = 4g HCl

ms=md+mH2O => mH2O = ms-md = 50-4 = 46g

n=m/M=4/36,5 = 0,1095 moli HCl

HCl in exces

94g.....2*36,5g...2*74,5g...18g
K2O + 2HCl => 2KCl + H2O
2,35g....x=1,825g...y=3,725g....z=0,45g

md final = (4-1,825) + 3,725  = 5,9g

ms final = 5,9+(46+0,45) = 52,35g sol

Cf=mdf/msf*100 = 5,9/52,35*100 = 11,27 %