pt: C+O2--->CO2(1)
CO2+Ca(OH)2--->CaCO3+H2O(2)
nCaCO3=0,5/100=0,005(mol)
Theo pt (1) và (2) : nC=nCO2(1)=nCO2(2)=nCaCO3=0,005(mol)
=>mC=0,005.12=0,06(g)
=>%mC=0,06/10.100%=0,6%
ptpu: C + O2= CO2
CO2 + Ca(OH)2=CaCO3+H2O
nCaCO3= 0.5:100=0.005 mol
=> nCO2=nC=nCaCO3= 0.005 mol (theo ptpư)
=> mC=0.005*12=0.06(g)
%C= (mC/mthép)*100%= (0.06/10)*100=0.6%