CaO+2HCl→CaCl2+H2O

nCaO=16,8/56=0,3mol

nHCl=0,1*2=0,2mol

Theo Pt nCaO/nHCl=1/2

Theo đề nCaO/nHCl=1,5/1

Vậy CaO dư

nCaO dư=0,3-0,2/2=0,2mol

nCaCl2=0,2/2=0,1mol

CM CaCl2=0,1/0,1=1M

b) mdd HCl=1,18*100=118g

mdd sau phản ứng=16,8+118=134,8g

mCaCl2=0,1*111=11,1g

C%CaCl2=11,1/134,8*100%=8,23%

mCa(OH)2 dư=0,2*74=14,8g

C%Ca(OH)2 dư=14,8/134,8*100%=10,98%

c) Ca(OH)2+H2SO4→CaSO4+H2O

     0,2       0,2

CaCl2+H2SO4→CaSO4+2HCl

0,1          0,1

nH2SO4=0,2+0,1=0,3mol

Vdd H2SO4=0,3/0,5=0,6l

Đáp án:

 \(\begin{array}{l}
a)\\
{C_M}Ca{(OH)_2} = 2M\\
{C_M}CaC{l_2} = 1M\\
b)\\
{C_\% }CaC{l_2} = 8,23\% \\
{C_\% }Ca{(OH)_2} = 10,98\% \\
c)\\
{V_{dd{H_2}S{O_4}}} = 0,6l
\end{array}\)

Giải thích các bước giải:

 \(\begin{array}{l}
a)\\
CaO + 2HCl \to CaC{l_2} + {H_2}O\\
{n_{CaO}} = \dfrac{{16,8}}{{56}} = 0,3\,mol\\
{n_{HCl}} = 0,1 \times 2 = 0,2\,mol\\
\text{ Lập tỉ lệ }:\dfrac{{{n_{CaO}}}}{1} > \dfrac{{{n_{HCl}}}}{2}(0,3 > 0,1)\\
 \Rightarrow \text{ CaO dư } \\
CaO + {H_2}O \to Ca{(OH)_2}\\
{n_{CaO}} \text{ dư }= 0,3 - 0,1 = 0,2\,mol\\
{n_{Ca{{(OH)}_2}}} = {n_{CaO}} \text{ dư }= 0,2\,mol\\
{n_{CaC{l_2}}} = \dfrac{{{n_{HCl}}}}{2} = \frac{{0,2}}{2} = 0,1\,mol\\
{C_M}Ca{(OH)_2} = \dfrac{{0,2}}{{0,1}} = 2M\\
{C_M}CaC{l_2} = \dfrac{{0,1}}{{0,1}} = 1M\\
b)\\
{m_{ddHCl}} = 100 \times 1,18 = 118g\\
{m_{ddspu}} = 16,8 + 118 = 134,8g\\
{C_\% }CaC{l_2} = \dfrac{{0,1 \times 111}}{{134,8}} \times 100\%  = 8,23\% \\
{C_\% }Ca{(OH)_2} = \dfrac{{0,2 \times 74}}{{134,8}} \times 100\%  = 10,98\% \\
c)\\
Ca{(OH)_2} + {H_2}S{O_4} \to CaS{O_4} + 2{H_2}O\\
CaC{l_2} + {H_2}S{O_4} \to CaS{O_4} + 2HCl\\
{n_{{H_2}S{O_4}}} = 0,2 + 0,1 = 0,3\,mol\\
{V_{dd{H_2}S{O_4}}} = \dfrac{{0,3}}{{0,5}} = 0,6l
\end{array}\)