CaCO3 + 2HCl ==> CaCl2 + H2O + CO2↑
nCaCO3=10/100=0,1 (mol)
a) Dựa vào pt, ta thấy:
nCO2=nCaCO3=0,1 (mol)
==> VCO2=0,1.22,4=2,24 (l)
b) nHCl=2nCaCO3=2.0,1=0,2 (mol)
=> mHCl=0,2.36,5=7,3 (g)
=> C%HCl= $\frac{7,3}{250}$ .100=2,92%
c) mdd sau phản ứng= mCaCO3 + mddHCl - mCO2= 10 + 250 - (0,1.44)=255,6 (g)
nCaCl2=nCaCO3=0,1 (mol) ==> mCaCl2=0,1.111=11,1 (g)
==> C%CaCl2= $\frac{11,1}{255,6}$ .100=4,34%