Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!!

Giải thích các bước giải:

a,

\(\begin{array}{l}
{m_{HCl}}{\rm{dd}} = D \times V = 200g\\
{n_{HCl}} = 0,04mol\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{NaOH}} = {n_{HCl}} = 0,04mol\\
 \to {V_{NaOH}} = \dfrac{{0,04}}{{0,1}} = 0,4l\\
{n_{NaCl}} = {n_{HCl}} = 0,04mol\\
 \to C{M_{NaCl}} = \dfrac{{0,04}}{{0,2 + 0,4}} = 0,067M
\end{array}\)

b,

\(\begin{array}{l}
{m_{HCl}}{\rm{dd}} = D \times V = 200g\\
{n_{HCl}} = 0,04mol\\
Ca{(OH)_2} + 2HCl \to CaC{l_2} + 2{H_2}O\\
{n_{Ca{{(OH)}_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,02mol\\
 \to {m_{Ca{{(OH)}_2}}} = 1,48g\\
 \to {m_{Ca{{(OH)}_2}}}{\rm{dd}} = \dfrac{{1,48 \times 100}}{5} = 29,6g\\
{n_{CaC{l_2}}} = {n_{Ca{{(OH)}_2}}} = 0,02mol\\
 \to {m_{CaC{l_2}}} = 2,22g\\
{m_{{\rm{dd}}}} = {m_{HCl}}{\rm{dd + }}{{\rm{m}}_{Ca{{(OH)}_2}}}{\rm{dd}} = 200 + 29,6 = 229,6g\\
 \to C{\% _{CaC{l_2}}} = \dfrac{{2,22}}{{229,6}} \times 100\%  = 0,97\% 
\end{array}\)

a)Đổi: $V_{HCl}$=200 ml=0,2 l

$n_{HCl}$=0,2.0,2=0,04 (mol)

HCl+NaOH→NaCl+$H_{2}$O

1----1--------1------1--------(mol)

0,04-0,04-----0,04---0,04----(mol)

$V_{NaOH}$=$\frac{0,04}{0,1}$=0,4 (l)

$V_{dd NaCl}$=0,2+0,4=0,6 (l)

$C_{M_{dd NaCl}}$=$\frac{0,04}{0,6}$=$\frac{1}{15}$ (M)

b)2HCl+$Ca(OH)_{2}$→$CaCl_{2}$+$H_{2}$O

--2------1---------------1------------1---------(mol)

--0,04---0,02------------0,02---------0,02-----(mol)

$m_{dd Ca(OH)_{2}}$=$\frac{0,02.74}{5%}$=29,6 (g)

$m_{dd CaCl_{2}}$=200.1+29,6=229,6 (g)

$C%_{dd CaCl_{2}}$=$\frac{0,02.111}{229,6}$.100%≈0,97%