化學計算題:計算碳酸鈣的純度。

碳酸鈣=CaCO3
氫氯酸=HCl
氫氧化鈉=NaOH

a)Reaction between NaOH and HCl:
NaOH + HCl = NaCl + H2O
1.Molar ratio of NaOH:HCl = 1:1
2.Number of mole of NaOH used to react HCl---->0.0290dm3X2.00M=0.0580mol
3.Number of mole of HCl used to react NaOH---->0.0580mol

b)Number of mole of 2.4MHCl in 50cm3---->0.0500dm3X2.400M=0.120mol

c)Reaction between CaCO3 and HCl:
CaCO3 + 2HCl = CaCl + H2O + CO2
1.1.Molar ratio of CaCO3:HCl = 1:2
2.Number of mole of HCl used to react CaCO3---->0.120mol-0.0580mol=0.0620mol
3.Number of mole of CaCO3 in the impure sample---->0.0620mol/2=0.0310mol

d)mass of CaCO3 in the impure sample---->0.0310X(40.1+12+16X3)=3.103g

e)percentage by mass of CaCO3 in the impure sample---->(3.103/3.6)X100%=86.2%

碳酸鈣的純度=86.2%