Răspunsuri

2014-01-19T22:20:35+02:00
CaCO3+H2SO4->CaSO4+CO2+H2O
banuiesc ca 20% inseamna impuritati
si avem asa
p/100=mp/mi => mp=p*mi/100=80*500/100=400 g CaCO3 pur
la 100g CaCO3.....1 mol CO2
la 400...................x
x=4 moli CO2 (conditii normale)
T=t+273.15=27.3+273.15=300.45 (am pus si cele 0.15 grade la zecimale)
PV=nRT => V=nRT/P=4*0.082*300.45/1=98.54 L (in cond cerute de problema)


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