a) cantitatea de substanță 2 mol Cl2
2 moli m g
2KI + Cl2 = 2KCl + I2↓
1 2x127
=> m = 2x2x127/1 = 508 g I2
b) volumul 56 L Cl2
56 L m g
2KI + Cl2 = 2KCl + I2↓
22,4 2x127
=> m = 56x2x127/22,4 = 635 g I2
c) masa de 355 g Cl2
355 g m g
2KI + Cl2 = 2KCl + I2↓
71 2x127
=> m = 355x2x127/71 = 1270 g I2