`a)`

`CaCO_3+2HCl→CaCl_2+H_2O+CO_2`

`b)`

`n_(CaCO_3)=60/100=0,6(mol)`

`n_(HCl)=0,6.2=1,2(mol)`

`m_(HCl)=1,2.36,5=43,8(g)`

`C%HCl=(43,8.100%)/400=10,95%`

`c)`

`m_(CaCl_2)=0,6.111=66,6(g)`

`d)`

`mdd=60+400-0,6.44=433,6(g)`

`C%CaCl_2=(66,6.100%)/(433,6)=15,36%`

Em tham khảo nha :

\(\begin{array}{l}
a)\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{n_{CaC{O_3}}} = \dfrac{{60}}{{100}} = 0,6mol\\
{n_{HCl}} = 2{n_{CaC{O_3}}} = 1,2mol\\
{m_{HCl}} = 1,2 \times 36,5 = 43,8g\\
C{\% _{HCl}} = \dfrac{{43,8}}{{400}} \times 100\%  = 10,95\% \\
c)\\
{n_{CaC{l_2}}} = {n_{CaC{O_3}}} = 0,6mol\\
{m_{CaC{l_2}}} = 0,6 \times 111 = 66,6g\\
d)\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,6mol\\
C{\% _{CaC{l_2}}} = \dfrac{{66,6}}{{60 + 400 - 0,6 \times 44}} \times 100\%  = 15,36\% 
\end{array}\)