\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)

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

        1          2               1          1

      0,25     0,5                          0,25

\(n_{H2}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,25.22,4=5,6\left(l\right)\)

\(n_{HCl}=\dfrac{0,25.2}{1}=0,5\left(mol\right)\)

250ml = 0,25l

\(C_{M_{ddHCl}}=\dfrac{0,5}{0,25}=2\left(M\right)\)

 Chúc bạn học tốt

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\\n_{HCl}=2n_{Fe}=0,4\left(mol\right)\end{matrix}\right.\)

a, Ta có: \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, \(V_{ddHCl}=\dfrac{0,4}{1,5}\approx0,267\left(l\right)\)

c, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

Bạn tham khảo nhé!

n_Mg = 0,1(mol)

PTHH: Mg + 2HCl --> MgCl₂ +H₂

n_Mg = n_MgCl2= n_H2 = 0,1(mol)

=> m_muối = 9,5(g) 

_VH2 = 2,24(l)

b) CM_HCl = 0,2/0,1=2(M)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\\ \left(mol\right)...0,2......\leftarrow...............0,2\\ m_{Zn}=0,2.65=13\left(g\right)\)

PTHH: Zn + 2HCl ^_____>  ZnCl₂ + H₂  (1)

Ta có: theo (1): n_{\(H_2\)(đktc)}=\(\dfrac{4.48}{22.4}\)=0.2 (mol)

theo (1): n_Zn = n_{\(H_2\)}= 0.2(mol)

=> m_Zn = 0.2 . 65 = 13(g)

Vậy giá trị m bằng 13(g)

$Fe_2O_3+3H_2\rightarrow 2Fe+3H_2O$

$Fe+2HCl\rightarrow FeCl_2+H_2$

Gọi số mol Fe2O3 là a

Ta có: $n_{H_2/(1)}=3a(mol);n_{H_2/(2)}=2a(mol)$

\(\Rightarrow\dfrac{V}{V'}=\dfrac{n_{H_2\left(1\right)}}{n_{H_2\left(2\right)}}=\dfrac{3}{2}\)

\(\left\{{}\begin{matrix}m_{Mg}=\dfrac{40.9}{100}=3,6\left(g\right)\\m_{Al}=9-3,6=5,4\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\\n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\end{matrix}\right.\)

PTHH:

Mg + 2HCl ---> MgCl₂ + H₂

0,15 ------------------------> 0,15

2Al + 6HCl ---> 2AlCl₃ + 3H₂

0,2 ---------------------------> 0,3

\(\rightarrow V_{H_2}=\left(0,15+0,3\right).22,4=10,08\left(l\right)\)

\(n_{H_2O}=\dfrac{6,48}{18}=0,36\left(mol\right)\)

PTHH: Fe_xO_y + yH₂ --t^o--> xFe + yH₂O

Theo pthh: \(n_{O\left(oxit\right)}=n_{H_2\left(pư\right)}=n_{H_2O}=0,36\left(mol\right)\)

\(\rightarrow m_{oxit}=15,12+16.0,36=20,88\left(g\right)\)

\(n_{Fe}=\dfrac{15,12}{56}=0,27\left(mol\right)\)

CTHH: Fe_xO_y

=> x : y = 0,27 :  0,36 = 3 : 4

=> CTHH: Fe₃O₄ (oxit sắt từ)

mMg = 40%x9 = 3,6(g) =>nMg=3,6:24 = 0,15 (mol)
=> mAl = 9-3,6 = 5,4(g) => nAl = 5,4:27 = 0,2 (mol)
pthh : 2Al+6HCl -> 2AlCl3+3H2
         0,2                            0,3
        Mg+2HCl -> MgCl2 +H2
        0,15                          0,15 
=> nH2 = 0,15 + 0,3 = 0,45 (mol) 
=> VH2 = 0,45.22,4 = 10,08 (L) 
mH2 = 0,45 . 2 = 0,9 (mol) 
áp dụng BLBTKL  ta có : 
mH2 + moxit sắt = mFe + mH2O 
=> moxit sắt =  20,7 (g) 
 

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

PTHH: Mg + 2HCl --> MgCl₂ + H₂

_____0,1--->0,2------->0,1---->0,1

=> m_HCl = 0,2.36,5 = 7,3(g)

b) m_MgCl2 = 0,1.95 = 9,5 (g)

c) V_H2 = 0,1.22,4 = 2,24(l)

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

PTHH: Mg + 2HCl --> MgCl₂ + H₂

_____0,1--->0,2------->0,1---->0,1

=> m_HCl = 0,2.36,5 = 7,3(g)

b) m_MgCl2 = 0,1.95 = 9,5 (g)

c) V_H2 = 0,1.22,4 = 2,24(l)