\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

a.

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

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

b.

\(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right),m_{CuO}=12-2,4=9,6\left(g\right)\)

c. 

\(m_{muối}=m_{CuCl_2}+m_{MgCl_2}=0,12.135+95.0,1=25,7\left(g\right)\)

\(a)n_{HCl}=0,2.1,5=0,3mol\\ CaO+2HCl\rightarrow CaCl_2+H_2O\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ \Rightarrow\left\{{}\begin{matrix}2n_{CaO}+2n_{CuO}=0,3\\56n_{CaO}+80n_{CuO}=10,8\end{matrix}\right.\\ \Rightarrow n_{CaO}=n_{CaCl_2}=0,05mol;n_{CuO}=n_{CuCl_2}=0,1mol\\ \%m_{CaO}=\dfrac{0,05.56}{10,8}\cdot100=25,93\%\\ \%m_{CuO}=100-25,93=74,07\%\\ b)C_{M_{CaCl_2}}=\dfrac{0,05}{0,2}=0,25M\\ C_{M_{CuCl_2}}=\dfrac{0,1}{0,2}=0,5M\)

\(CaO+2HCl\rightarrow CaCl_2+H_2O\)

x            2x            x           x 

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

y             2y            y          y 

\(\left\{{}\begin{matrix}56x+80y=10,8\\2x+2y=0,2.1,5=0,3\end{matrix}\right.\)

\(\Rightarrow x=0,05;y=0,1\)

\(a,\%m_{CaO}=0,05.56:10,8.100\%=25,93\left(\%\right)\)

\(\%m_{CuO}=100\%-25,93\%=74,07\%\)

\(b,C_{M\left(CaCl_2\right)}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)

\(C_{M\left(CuCl_2\right)}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

a) PTHH: Fe + 2HCl ===> FeCl₂ + H₂

MgO + 2HCl ===> MgCl₂ + H₂O

b) Ta có: n_H2 =\(\frac{2,24}{22,4}=0,1\left(mol\right)\)

=> n_Fe = n_H2 = 0,1 (mol)

=> m_Fe = 0,1 x 56 = 5,6 gam

=> m_MgO = 22,4 - 5,6 = 156,8 gam

c. %_MgO = \(\frac{16,8}{22,4}.100\%=75\%\)

=> %_Fe = 100% - 75% = 25%

nH2=0,1(mol)

PTHH: Mg + 2 HCl -> MgCl2 + H2

0,1__________0,2___________0,1(mol)

MgO + 2 HCl -> MgCl2 + H2O

0,05____0,1___0,05(mol)

mMg=0,1. 24= 2,4(g) -> mMgO=4,4-2,4= 2(g) -> nMgO=0,05((mol)

b) %mMg= (2,4/4,4).100=54,545%

=> %mMgO=45,455%

c) nHCl=0,3(mol) -> mHCl=0,3.36,5=10,95(g)

=> mddHCl=(10,95.100)/7,3=150(g)

a) CaO+2HCl--->CaCl2+H2O

x-----------2x----------x

b) CuO+2HCl---->CuCl2+H2O

y----------2y

b) n HCl=0,4.0,1=0,04(mol)

Theo bài ta có hpt

\(\left\{{}\begin{matrix}56x+80y=1,36\\2x+2y=0,04\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,01\\y=0,01\end{matrix}\right.\)

%m CaO=0,01.56/1,36.100%=41,18%

%m CuO=100%-41,18=58,82%

c) m muối=0,01.111+0,01.135=2,46(g)

Chúc bạn học tốt

a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

             1         2                1          1

            0,3     0,6                            0,3

           \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

               1           2               1           1

              0,1        0,2

b) \(n_{Mg}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(m_{Mg}=0,3.24=7,2\left(g\right)\)

\(m_{MgO}=11,2-7,2=4\left(g\right)\)

c) ⁰/₀Mg = \(\dfrac{7,2.100}{11,2}=64,29\)⁰/₀

     ⁰/₀MgO = \(\dfrac{4.100}{11,2}=35,71\)⁰/₀

d) Có : \(m_{MgO}=4\left(g\right)\)

\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,6+0,2=0,8\left(mol\right)\)

\(m_{HCl}=0,8.36,5=29,2\left(g\right)\)

\(C_{ddHCl}=\dfrac{29,2.100}{200}=14,6\)⁰/₀

 Chúc bạn học tốt

cảm ơn nhìu nhìu

\(n_{FeO}=a\left(mol\right),n_{CuO}=b\left(mol\right)\)

\(m_{hh}=72a+80b=19.2\left(g\right)\left(1\right)\)

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(n_{H_2SO_4}=a+b=0.25\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.1,b=0.15\)

\(m_{FeO}=0.1\cdot72=7.2\left(g\right)\)

\(m_{CuO}=12\left(g\right)\)

\(C_{M_{FeSO_4}}=\dfrac{0.1}{0.25}=0.4\left(M\right)\)

\(C_{M_{CuSO_4}}=\dfrac{0.15}{0.25}=0.6\left(M\right)\)