irinafilipoiu
CaC2 + CO2 -> CO + CaO
Daca am 10g de p=82% CaC2, atunci m=820/100=8,2g M=64g/mol, deci n=8,2/64=0,128125moli, prin hidroliza CaC2 + H2O-> C2H2 + Ca(OH)2, Ca(OH2) + CO2-> CaCO3 + H2O, din reactii rezulta ca nCO2=0,128125 moli, C6H12O6 -> 2C2H5OH + 2CO2, deci n glucoza=0,128125/2=0,0640625 moli, m=11,53125g, V=0,025L, iar Cmolar=m/V=461,25g/L (vezi ca la Iasi mai si aproximeaza daca la asta te referi)