2Mg + O₂ → 2MgO
a) M Mg = 24 g
cantitatea= 5 moli
24 * 5 = 120 g Mg necesare.
b)M O₂= 16 g/mol
2*24/120 = 16/x
x= 120*16/48
x= 1920/48
x= 40 g O₂ necesare.
1 mol O₂.......16 g
x moli O₂......40 g
1/x=16/40
x= 40/16
x= 2,5 moli O₂
Volumul:
1 mol O₂...... 22,4 l
2,5 moli O₂......x l
x = 2,5 * 22,4
x= 56 litri O₂