Question

để trung hòa 125ml dd Ca(OH)2 cần dùng 182,5g HCl 10%. tính \(C_{M_{Ca\left(OH\right)_2}}\)

Answer 1

Ca(OH)2+2HCl----->CaCl2 +2H2O

m\(_{HCl}=\frac{182,5.10}{100}=18,25\left(g\right)\)

n\(_{HCl}=\frac{18,25}{36,5}=0,5\left(mol\right)\)

Theo pthh

n\(_{Ca\left(OH\right)2}=\frac{1}{2}n_{HCl}=0,25\left(mol\right)\)

C\(_{M\left(Ca\left(OH\right)2\right)}=\frac{0,25}{0,125}=2\left(M\right)\)

Answer 2

mHCl =18,25 g =>mHCl = 0,5(mol)

PTHH: Ca(OH)2 +2HCl--->CaCL2+2H2O

=>nCa(OH)2 = 0,25 (mol)

=>CM = 0,25 / 0,125 = 2(M)