M,MgO= (24gMg+16gO)/mol= 40g/mol-------> niu(n)=m/M=>120g/40g/mol=3molMgO
M,MgCO3=(24gMg+12gC+3X16gO)/mol=84g/mol
M,CO2= (12gC+2x16gO)/mol= 44g/mol
carbonatul de Mg se descompune:
1mol 1mol 1mol
MgCO3----->Mg0+ CO2
x..................3............y
x=3mol MgCO3--------> m= nxM m =3molx.84g/mol=..........calculeaza !
y =3molCO2-----------> m=3molx44g/mol= ....calculeaza !!