Question

De egalat si de verificat Legea conservării masei pentru:
7) Mg + O2 = MgO
2) Na2O + H₂O = NaOH
3)MgO + H2O Mg(OH)2
Mg:24
O:16
H:1
Na:23​

Answer 1

Răspuns:

Explicație:

2Mg + O2= 2MgO

2.24 g   + 32g   = 2.40g

80g = 80g

Na2O + H2O= 2 NaOH

62g + 18g = 2.40g

80g = 80g

MgO + H2O= Mg(OH)2

40g + 18g = 58g

58g=48g

Answer 2

Răspuns:

2 Mg+ O2=> 2 MgO

MMg=24g/mol

MO2=2×16=32g/mol

MMgO=24+16=40g/mol

reactanti 2×24+32=80g

produși de reactie 2×40=80g

2) Na2O+ H2O => 2NaOH

MNa2O=2×23+16=62g/mol

MH2O =2×1+16=18g/mol

MNaOH =23+16+1=40g/mol

reactanti 62+18=80g

produși de reactie 2×40=80g

3. MgO + H2O => Mg(OH)2

MMgO=24+16=40g/mol

MH2O=2×1+16=18g/mol

MMg(OH)2=24+2×16+2×1=58g/mol

reactanti 40+18=58g

produși de reactie 58g