Question

1.Cati moli de oxigen sunt necesari pentru a obtine 306 g oxid de aluminiu? 2. 12g Mg reactioneaza cu HCl.Calculeaza masa de clorura de magneziu obtinuta si numarul de moli H care se obtin.

Answer 1

1. MAl₂O₃ = 2AAl+3AO=2*27+3*16=102g/mol

n=m/M=306/102=3 moli Al₂O₃

4Al + 3O₂ ⇒ 2Al₂O₃

3 moli O₂....................2 moli Al₂O₃
x moli O₂....................3 moli Al₂O₃

x=3*3/2=Calculeaza !!

2. MMg=24g/mol

MMgCl₂=AMg+2ACl=24+2*35,5=95g/mol

n=m/M=12/24=0,5 moli Mg

Mg + 2HCl ⇒ MgCl₂ + H₂↑

24g Mg........................95g MgCl₂
12g Mg.........................xg MgCl₂

x=95*12/24= Calculeaza !!

1 mol Mg............................1 mol H₂
0,5 moli Mg.........................0,5 moli H₂

M-masa molara,g/mol

A-masa atomica

m-masa,g

n-nr de moli