Question

Ajutatima va rog este evaluare ultimul punct.

Answer 1

M CaCO3=100 g/mol

M CaCl2=111 g/mol

M H2O=18 g/mol

Masa pură de CaCO3:

0,8×200=160 g=>n

CaCO3=CaO+CO2

n CaO=1,6 moli

2HCI+CaO=CaCl2+H2O

nCaCl2=1,6 moli

nH2O=1,6 moli

nCaCl2=111×1,6=177,6 g

mH2O=18×1,6=28,8 g

Answer 2

CaCO3 = CaO + CO2
CaO + 2 HCl = CaCl2 + H2O
M CaCO3 = 100 g/mol
M CaO = 56 g/mol
M HCl = 36,5 g/mol
M CaCl2 = 111 g/mol
M H2O = 18 g/mol
puritatea = mpură/m impură * 100
m pură = 200*80/100 = 160 g CaCO3
100 g CaCO3.........56 g CaO
160 g CaCO3..............x
x = 89,6 g CaO
56 g CaO.......111 g CaCl2.........18 g H2O
89,6 g CaO................y...................z
y = 177,6 g CaCl2
n = m/M = 177,6/111 = 1,6 moli CaCl2
z = 28,8 g H2O
n = m/M = 28,8/18 = 1,6 moli H2O