\(n_{Fe_3O_4}=0,01\left(mol\right)\\ Fe_3O_4+8HCl\rightarrow2FeCl_3+FeCl_2+4H_2O\\ n_{HCl}=1.1=1\left(mol\right)\\ V\text{ì}:\dfrac{0,01}{1}< \dfrac{0,1}{8}\Rightarrow HCl\text{dư}\\ \Rightarrow\text{dd}X:FeCl_2,FeCl_3,HCl\left(d\text{ư}\right)\\ n_{FeCl_2}=n_{Fe_3O_4}=0,01\left(mol\right)\\ n_{FeCl_3}=0,01.2=0,02\left(mol\right)\\ n_{HCl\left(d\text{ư}\right)}=1-0,01.8=0,92\left(mol\right)\\ V_{\text{dd}X}=V_{\text{dd}HCl}=1\left(l\right)\\ C_{M\text{dd}HCl\left(d\text{ư}\right)}=\dfrac{0,92}{1}=0,92\left(M\right)\\ C_{M\text{dd}FeCl_2}=\dfrac{0,01}{1}=0,01\left(M\right)\\ C_{M\text{dd}FeCl_3}=\dfrac{0,02}{1}=0,02\left(M\right)\)

a)

$Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O$

b)

n HCl = 0,4.1,5 = 0,6(mol)

n Al2O3 = 1/6 n HCl = 0,1(mol) => m = 0,1.102 = 10,2(gam)

n AlCl3 = 1/3 n HCl = 0,2(mol) => CM AlCl3 = 0,2/0,4 = 0,5M

nHCl = 0.4*1.5 = 0.6 (mol) 

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

nAlCl3 = 0.6*2/6 = 0.2 (mol) 

mAlCl3 = 0.2*133.5 = 26.7 (g) 

CM AlCl3 = 0.2/0.4 = 0.5 (M) 

$n_{CuO} = \dfrac{8}{80} = 0,1(mol) ; n_{HCl} = 0,15.2 = 0,3(mol)$
$CuO + 2HCl \to CuCl_2 + H_2O$
Ta thấy : 

$n_{CuO} : 1 < n_{HCl} : 2$ nên HCl dư

$n_{CuCl_2} = n_{CuO} = 0,1(mol)$
$n_{HCl\ pư} = 2n_{CuO} = 0,2(mol) \Rightarrow n_{HCl\ dư} = 0,3 - 0,2 = 0,1(mol)$
$C_{M_{CuCl_2}} = \dfrac{0,1}{0,15} = 0,67M$
$C_{M_{HCl}} = \dfrac{0,1}{0,15} = 0,67M$

1.

n_Al=\(\dfrac{5,4}{27}\)=0,2 mol

m_HCl=\(\dfrac{175.14,6}{100}\)=25,55g

n_HCl=\(\dfrac{25,55}{36,5}\)=0,7

                     2Al + 6HCl → 2AlCl₃ + 3H₂↑

n trước pứ    0,2     0,7  

n pứ              0,2 →0,6  →    0,2  →  0,3  mol

n sau pứ       hết   dư 0,1

Sau pứ HCl dư.

m_HCl (dư)= 36,5.0,1=3,65g

m_{các chất sau pư}= 5,4 +175 - 0,3.2= 179,8g

m_AlCl3= 133,5.0,2=26,7g

C%_ddHCl (dư)= \(\dfrac{3,65.100}{179,8}=2,03%\)%

C%_ddAlCl3 = \(\dfrac{26,7.100}{179,8}\)= 14,85%

2.

           200ml= 0,2l

m_Mg= \(\dfrac{4,2}{24}=0,175mol\)

    Mg + 2HCl → MgCl₂ + H₂↑

0,175→ 0,35 → 0,175→0,175 mol

a) V_H2= 0,175.22,4=3,92l.

b)C%_dHCl= \(\dfrac{0,35}{0,2}=1,75\)M

nNaOH = 0,1 mol

nH2SO4 = 0,1 mol

PT: 2NaOH + H2SO4 -> Na2SO4 + H2O

=> H2SO4 dư: 0,1 - 0,05= 0,05 (mol)

=> mH2SO4 dư = n. M = 0,05 . 98 = 4,9 g

a,\(2NaOH+H2SO4->Na2SO4+2H2O\)

b,theo pthh 

PTHH:\(2NaOH+H2SO4->Na2SO4+2H2O\)

theo pthh:\(2\)..................1...........(mol)

theo bài: \(\dfrac{100}{1000}\)............\(\dfrac{9,8}{98}........\)(mol)

\(=>\dfrac{0,1}{2}< \dfrac{0,1}{1}\)=>H2SO4 dư

c,theo pthh \(=>nNA2SO4=\dfrac{1}{2}nNaOH=0,05mol\)

\(=>mNa2SO4=142.0,05=7,1g\)

Bổ sung: \(D_{HCl}=1,18\left(g/ml\right)\)

a) PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

b) Ta có: \(\left\{{}\begin{matrix}n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\\n_{HCl}=\dfrac{100\cdot1,18\cdot20\%}{36,5}=\dfrac{236}{365}\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{\dfrac{236}{365}}{2}\) \(\Rightarrow\) HCl còn dư, MgO p/ứ hết

\(\Rightarrow n_{MgCl_2}=0,05\left(mol\right)\) \(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)

Câu 1:

a) CaCO₃ + 2HCl --> CaCl₂ + CO₂ + H₂O

CaO + 2HCl --> CaCl₂ + H₂O

b) 

\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: CaCO₃ + 2HCl --> CaCl₂ + CO₂ + H₂O

_____0,4<---------0,8<-----------------0,4

=> m_CaCO3 = 0,4.100 = 40(g)

=> m_CaO = 62,4 - 40 = 22,4 (g)

c) \(n_{CaO}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

CaO + 2HCl --> CaCl₂ + H₂O

_0,4-->0,8

=> n_HCl = 0,8 + 0,8 = 1,6(mol)

=> \(C_{M\left(HCl\right)}=\dfrac{1,6}{0,25}=6,4M\)

Câu 1:

\(a,CaO+2HCl\to CaCl_2+H_2O\\ CaCO_3+2HCl\to CaCl_2+H_2O+CO_2\uparrow\\ b,n_{CO_2}=\dfrac{8,96}{22,4}=0,4(mol)\\ \Rightarrow n_{CaCO_3}=0,4(mol)\\ \Rightarrow m_{CaCO_3}=0,4.100=40(g)\\ \Rightarrow m_{CaO}=62,4-40=22,4(g)\\ c,n_{CaO}=\dfrac{22,4}{56}=0,4(mol)\\ \Rightarrow \Sigma n_{HCl}=0,4.2+0,4.2=1,6(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{1,6}{0,25}=6,4M\)

Câu 2: Đề thiếu

\(a.Fe+CuSO_4\rightarrow FeSO_4+Cu\\ b.n_{Cu}=\dfrac{3,2}{64}=0,05mol\\ n_{CuSO_4}=\dfrac{100.1,12.10}{100}:160=0,7mol\\ \Rightarrow\dfrac{0,05}{1}< \dfrac{0,07}{1}\Rightarrow CuSO_4.dư\\ Fe+CuSO_4\rightarrow FeSO_4+Cu\)

0,05  0,05          0,05        0,05 (mol)

\(C_M\) \(_{FeSO_4}=\dfrac{0,05}{0,1}=0,5M\)

\(C_M\) \(_{CuSO_4}=\dfrac{0,07-0,05}{0,1}=0,2M\)