Question

ajutați-mă va rog la aceasta problema​

Answer 1

Răspuns:

Cu(OH)2 + H2CO3 → CuCO3 + 2H2O

a) MCu(OH)2=64+2x16+2x1=98g/mol

MH2CO3=2x1+12+3x16=62g/mol

c)

98g.................62g

Cu(OH)2 + H2CO3 → CuCO3 + 2H2O

xg....................31g

x=31x98/62=49g Cu(OH)2

d)

1mol.....................................1mol

Cu(OH)2 + H2CO3 → CuCO3 + 2H2O

0,5moli............................ymoli

y=0,5x1/1=0,5moli CuCO3

n=m/M=49g/ 98=0,5moliCu(OH)2

e)

md=31g H2CO3

C=10%

ms=?

ms=mdx100/C=31x100/10=310gH2CO3

Explicație: