Bài 5: nCaCO3=20/100=0,2(mol)
PTHH: Ca(OH)2 + CO2 -> CaCO3 + H2O
0,2______________0,2__0,2(mol)
PTHH: C6H12O6 --men rượu, 30-35 độ C---> 2 C2H5OH + 2 CO2
nC6H12O6(LT)= nCO2/2= 0,2/2=0,1(mol)
=> nC6H12O6(TT)= 0,1/80%= 0,125(mol)
=> mC6H12O6(TT)= 0,125.180= 22,5(g)
=> m=22,5(g)
