20%-impuritati=> puritate de 80%. 80/100*300=240g CaCO3 pur (care reactioneaza) CaCO3--->CaO+O2. Gazul este O2. Miu CaCO3=40+12+16*3=52+48=100g/mol miu O2=32 g/mol. 240/100=x/32=>x=76.8 g O=2.4 moli O n(nr. moli)=m/miu=V/Vm=N/Na=>V=n*Vm=2.4*22.4=53.76 L O