a)

2Al  +   6HCl  →  2AlCl₃  +  3H₂

b) nAl  = 5,4 : 27 = 0,2 mol

Theo tỉ lệ phản ứng => nH₂ = 0,3 mol <=> VH₂ = 0,3.22,4 = 6,72 lít.

c) nAlCl₃ = nAl = 0,2 mol

=> mAlCl₃ = 0,2. 133,5 = 26,7 gam.

d) nHCl cần dùng = 3nAl = 0,6 mol

=> mHCl = 0,6.36,5 = 21,9 gam

<=> mdd HCl cần dùng = \(\dfrac{21,9}{3,65\%}\) = 600 gam

a.b.c.\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,1                         0,1      0,15   ( mol )

\(V_{H_2}=0,15.22,4=3,36l\)

\(m_{AlCl_3}=0,1.133,5=13,35g\)

d.\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)

                  0,15              0,1              ( mol )

\(m_{Fe}=0,1.56=5,6g\)

a)\(PTHH:Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\)

b) Số mol của magie là:

\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}=\dfrac{3,6}{24}=0,15mol\)

\(PTHH:Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\)

tỉ lệ       :1          2           1               1      mol

số mol  :0,15     0,3        0,15          0,15 mol

Thể tích khí hiđro là:

\(V_{H_2}=n_{H_2}.22,4=0,15.22,4=3,36\left(l\right)\)

c) Khối lượng muối tạo thành là:

\(m_{MgCl_2}=n_{MgCl_2}.M_{MgCl_2}=0,15.95=14,25\left(g\right)\)

d) Khối lượng axit đã dùng là:

\(m_{HCl}=n_{HCl}.M_{HCl}=0,3.36,5=10,95\left(g\right)\)

a)

$Fe + 2HCl \to FeCl_2 + H_2$
n H2 = n Fe = 11,2/56 = 0,2(mol)

V H2 = 0,2.22,4 = 4,48(lít)

b)

n HCl = 2n Fe = 0,2.2 = 0,4(mol)

=> CM HCl = 0,4/0,4 = 1M

c)

$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

Ta thấy :

n CuO = 64/80 = 0,8 > n H2 = 0,2 nên CuO dư

Theo PTHH :

n CuO pư = n Cu = n H2 = 0,2(mol)

n Cu dư = 0,8 - 0,2 = 0,6(mol)

Vậy : 

%m Cu = 0,2.64/(0,2.64 + 0,6.80)   .100% = 21,05%
%m CuO = 100% -21,05% = 78,95%

a) 2Al + 6HCl → 2AlCl3 + 3H2

b) n Al= 8,1/27 =0,3(mol)

n H2 = 3/2 n Al = 0,45(mol)

V H2 = 0,45.22,4= 10,08 lít

c) n AlCl3 = n Al = 0,3 mol

m AlCl3 = 0,3.133,5 = 40,05(gam)

d) n HCl = 2n H2 = 0,9(mol)

=> V dd HCl = 0,9/0,5 = 1,8(lít)

a)

$Mg + 2HCl \to MgCl_2 + H_2$

Theo PTHH : 

$n_{MgCl_2} = n_{H_2} = \dfrac{5,6}{22,4} = 0,25(mol)$
$m_{MgCl_2} = 0,25.95 = 23,75(gam)$

b)

$n_{HCl} = 2n_{H_2} = 0,5(mol)$
$m_{dd\ HCl} = \dfrac{0,5.36,5}{7,3\%} = 250(gam)$

c)

$2K + 2H_2O \to 2KOH + H_2$
$n_K = 2n_{H_2} = 0,5(mol)$
$m_K = 0,5.39 = 19,5(gam)$

`Zn + 2HCl -> ZnCl_2 + H_2`

`0,2`  `0,4`                 `0,2`     `0,2`      `(mol)`

`n_[Zn]=13/65=0,2(mol)`

`a)V_[H_2]=0,2.22,4=4,48(l)`

`b)C%_[HCl]=[0,4.36,5]/100 . 100 =14,6%`

`c)C%_[ZnCl_2]=[0,2.136]/[13+100-0,2.2].100~~24,16%`

`d)`

`H_2 + CuO` $\xrightarrow{t^o}$ `Cu + H_2 O`

`0,1`     `0,1`          `0,1`                    `(mol)`

`n_[CuO]=8/80=0,1(mol)`

Ta có:`[0,2]/1 > [0,1]/1`

  `=>H_2` dư, `CuO` hết

`=>m_[Cu]=0,1.64=6,4(g)`

\(n_{Zn}=\dfrac{13}{65}=0,2mol\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

0,2      0,2                0,2       0,2

a)\(V_{H_2}=0,2\cdot22,4=4,48l\)

b)\(m_{H_2SO_4}=0,2\cdot98=19,6g\)

   \(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{19,6}{100}\cdot100\%=19,6\%\)

c)\(m_{ZnSO_4}=0,2\cdot161=32,2g\)

   \(m_{ddZnSO_4}=13+100-0,2\cdot2=112,6g\)

   \(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{32,2}{112,6}\cdot100\%=28,6\%\)

d)\(n_{CuO}=\dfrac{8}{80}=0,1mol\)

   \(CuO+H_2\rightarrow Cu+H_2O\)

   0,1        0,2     0,1

   \(m_{Cu}=0,1\cdot64=6,4g\)

\(n_{Fe}=\dfrac{16.8}{56}=0.3\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0.3....................0.3.........0.3\)

\(m_{FeCl_2}=0.3\cdot127=38.1\left(g\right)\)

\(V_{H_2}=0.6\cdot22.4=6.72\left(l\right)\)

\(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)

\(.......0.3...0.3\)

\(m_{Cu}=0.3\cdot64=19.2\left(g\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

a, Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

Theo PT: \(n_{FeCl_2}=n_{Fe}=0,3\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=0,3.127=38,1\left(g\right)\)

b, Theo PT: \(n_{H_2}=n_{Fe}=0,3\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

c, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Theo PT: \(n_{Cu}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{Cu}=0,3.64=19,2\left(g\right)\)

Bạn tham khảo nhé!