200ml = 0,2l

\(n_{HCl}=1.0,2=0,2\left(mol\right)\)

a) Pt : \(CaO+2HCl\rightarrow CaCl_2+H_2O|\)

             1            2             1            1

           0,1          0,2         0,1

b) \(n_{CaO}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{CaO}=0,1.40=4\left(g\right)\)

c) \(n_{CaCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{CaCl2}=0,1.111=11,1\left(g\right)\)

d) \(C_{M_{CaCl2}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

 Chúc bạn học tốt

a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

b) Ta có: \(n_{HCl}=0,05\cdot0,3=0,015\left(mol\right)\)

\(\Rightarrow n_{Fe}=0,0075\left(mol\right)\) \(\Rightarrow m_{Fe}=0,0075\cdot56=0,42\left(g\right)\)

c+d) Theo PTHH: \(n_{FeCl_2}=n_{H_2}=n_{Fe}=0,0075\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,0075\cdot22,4=0,168\left(l\right)\\m_{FeCl_2}=0,0075\cdot127=0,9525\left(g\right)\end{matrix}\right.\)

Bài 9 : 

\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

0,05--->0,1-------->0,05 

a) \(C_{MddHCl}=\dfrac{0,1}{0,1}=1\left(M\right)\)

b) \(m_{CuCl2}=0,05.135=6,75\left(g\right)\)

c) \(C_{MCuCl2}=\dfrac{0,05}{0,1}0,5\left(M\right)\)

Câu 10 : 

\(n_{FeO}=\dfrac{3,6}{72}=0,05\left(mol\right)\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

0,05-->0,1------->0,05

\(m_{ddHCl}=\dfrac{0,1.36,5}{10\%}100\%=36,5\left(g\right)\)

\(m_{ddspu}=3,6+36,5=40,1\left(g\right)\)

\(C\%_{FeCl2}=\dfrac{0,05.127}{40,1}.100\%=15,84\%\)

Câu 10 :

\(n_{FeO}=\dfrac{3,6}{72}=0,05\left(mol\right)\)

PTHH :

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

0,05      0,1           0,05       0,05

\(m_{ddHCl}=\dfrac{0,1.36,5.100}{10}=36,5\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{0,05.127}{\left(3,6+36,5\right)}.100\%=15,84\%\)