Câu 1:

m ct HCl=$\frac{200.7,3}{100}$=14,6 g

n HCl=$\frac{14,6}{36,5}$=0,4 mol

CaCO3+2HCl→CaCl2+CO2↑+H2O

 0,2    ←0,4  →   0,2       0,2              mol

mdd sau pứ=m CaCO3+mdd HCl-m CO2

                    =0,2.100+200-0,2.44 =211,2 g

m ct CaCl2=0,2.111=22,2 g

⇒C% CaCl2=$\frac{22,2}{211,2}$.100≈10,51 %

Câu 2:

a.

n Na2O=$\frac{31}{62}$=0,5 mol

Na2O+H2O→2NaOH

 0,5→                    1                mol

m NaOH=1.40=40 g

b.

Vì D H2O=1 g/ml

⇒mdd H2O=200.1=200 g

mdd sau=m Na2O+m H2O

               =31+200 =231 g

⇒C% NaOH=$\frac{40}{231}$.100≈17,316 %

------------------Nguyễn Hoạt--------------------

Em tham khảo nha:

\(\begin{array}{l}
1)\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{m_{HCl}} = \dfrac{{{m_{{\rm{dd}}}} \times {C_\% }}}{{100}} = \dfrac{{200 \times 7,3}}{{100}} = 14,6(g)\\
{n_{HCl}} = \dfrac{m}{M} = \dfrac{{14,6}}{{36,5}} = 0,4\,mol\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = \dfrac{{{n_{HCl}}}}{2} = \dfrac{{0,4}}{2} = 0,2\,mol\\
{m_{{\rm{dd}}spu}} = {m_{CaC{O_3}}} + {m_{{\rm{dd}}HCl}} - {m_{C{O_2}}} = 0,2 \times 100 + 200 - 0,2 \times 44 = 211,2g(g)\\
{n_{CaC{l_2}}} = {n_{C{O_2}}} = 0,2\,mol\\
{C_\% }(CaC{l_2}) = \dfrac{{0,2 \times 111}}{{211,2}} \times 100\%  = 10,51\% \\
2)\\
a)\\
N{a_2}O + {H_2}O \to 2NaOH\\
{n_{N{a_2}O}} = \dfrac{m}{M} = \dfrac{{31}}{{62}} = 0,5\,mol\\
 \Rightarrow {n_{NaOH}} = 2{n_{N{a_2}O}} = 0,5 \times 2 = 1\,mol\\
{m_{NaOH}} = n \times M = 1 \times 40 = 40g\\
b)\\
{m_{{H_2}O}} = V \times D = 200 \times 1 = 200g\\
{m_{{\rm{dd}}NaOH}} = 31 + 200 = 231(g)\\
{C_\% }(NaOH) = \dfrac{m}{{{m_{{\rm{dd}}}}}} \times 100\%  = \dfrac{{40}}{{231}} \times 100\%  = 17,32\% 
\end{array}\)