Question

4,8 g de mg se supun arderii. cati moli de oxid bazic se obtin? ce cantiate de oxigen avem nevoie?

Answer 1

Răspuns:

calculam in grame:

MMg=24g/mol

MO2=2×16=32g/mol

MMgO=24+16=40g/mol

2×24g....32g....2×40g

2Mg  +  O2  → 2MgO

4,8g.......xg.......yg

x=4,8×32/2×24=calculeaza

y=4,8×2×40/2×24=calculeaza

calculam in moli

n=m/M=4,8/24=0,2moli Mg

2moli......1mol.......2moli

2Mg  +  O2    →   2MgO

0,2moli....zmoli......tmoli

z=0,2×1/2= calculeaza

t=0,2×2/2 = calculeaza

Explicație: