100% - 20% impuritati = 80% puritate
100% ...................... 20 g MgCO3 impur
80% ........................ m = 16 g MgCO3 pur
16 g m g V L
MgCO3 + 2HNO3 --> Mg(NO3)2 + H2O + CO2
84 148 22,4
=> m = 16x148/84 = 28,20 g sare formata
=> V = 16x22,4/84 = 4,30 L gaz degajat