mCO2=?g
n= 3moli H₂CO₃
CO₂ + H₂O⇄ H₂CO₃
₋ ₋
subliniem dioxid de carbon si acid carbonic,apoi calculam masele moleculare
MCO₂=12+(2*16)=44 g
MH₂CO₃=( 2*1)+ 12+(3*16)=2+12+48=62g
transformammoli in grame cu relatia n= m/M→m=n*M
m=3*62=186g acid carbonic
44g CO₂.........................62g H₂CO₃
x ........................................186gH₂CO₃
x= 44*186/62 =132g CO₂
rezultatul in moli n=m/M
n=132/44=3 moli CO₂
