Question

248 g CuCO3 se descompun termic,calculati cantitatile produsilor de reactie rezultati.
Va implor ajutati-ma ☺​

Answer 1

Ecuația reacției de descompunere:

2 moli                       x                 y

CuCO₃      ----->      CuO     +     CO₂

1 mol                      1 mol          1 mol

Rezolvare:

$m_{CuCO_3}=248\ g\\M_{CuCO_3}=65+12+3\cdot16=125\ g/mol\\n=\frac{m}{M}=\frac{248}{125}=1.984\ moli\ (\approx2\ moli)\\\\n_{CuO}=x=1.984\ moli\\M_{CuO}=65+16=81\ g/mol\\n=\frac{m}{M}\implies m=n\cdot M=1.984\cdot81=160.7\ g\ CuO\\\\n_{CO_2}=y=1.984\ moli\\M_{CO_2}=12+2\cdot\ 16=12+32=44\ g/mol\\n=\frac{m}{M}\implies m=n\cdot M=1.984\cdot 44=87.3\ g\ CO_2\\\\m_{CuO}+m_{CO_2}=160.7+87.3=248\ g\ (m_{CuCO_3})$

Answer 2

CuCO₃=CuO+CO₂

MCuCO3=64+12+3*16=124u.a.m

MCuO=64+16=76 u.a.m

MCO₂=12+2*16=44 u.a.m

124 gCuCO3.......76gCuO......44gCO2

248 g.................................x                y

x=152 g CuO    y=88g