Question

O cantitate de 3 moli HCl reactioneaza cu calcar de puritate 70%.Sa se calculeze:
A. Cantitatea de calcar imput
B.Cantitatea de sare rezultata din reactie

Answer 1

-ecuatia reactiei
2mol........1mol............1mol
2HCl   + CaCO3---> CaCl2 + CO2 + H2O
3mol........x................y

y=m,CaCl2=3/2molx111g/mol=..........................calculeaza !!!
x= m,CaCO3pur= 3/2molx100g/mol= 150g CaCO3 PUR

Masa de carbonat impur se afla din puritate
p= m,purx100/m,impur
70/100= 150/m,impur--------> m,impur= 150x100/70=..............calculeaza !!!