Question

Sa se calculeze volumul gazului rezultat prin reactia a 400g CaCO3 cu 20% impuritati ce reactioneaza cu 200g solutie HCl de concentratie 30%. Care este masa substantei in exces?

Answer 1

0,8*400=320g CaCO3 pur

n=m/M=320/100=3,2 moli CaCO3

C=md*100/ms=>md=30*200/100=60g HCl

n=m/M=60/36,5=1,6438 moli HCl

CaCO3 + 2HCl => CaCl2 + CO2 + H2O

1 mol CaCO3........2 moli HCl
x=0,821 moli.................1,6438 HCl

n exces CaCO3 = 3,2-0,821=2,378

m exces = 2,378*100=237,8g CaCO3

nHCl=nCO2=1,6438 moli

VCO2=1,6438*22,4=36,821L CO2