a) PTHH: Fe + 2HCl --> FeCl₂ + H₂

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            0,2-->0,4------>0,2-->0,2

=> \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c) \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

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

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1\left(mol\right)\\ m_{muối}=m_{ZnCl_2}=0,1.136=13,6\left(g\right)\\ V_{khí\left(đktc\right)}=V_{H_2\left(đkc\right)}=0,1.24,79=2,479\left(l\right)\\ c,n_{CuO}=\dfrac{7,6}{80}=0,095\left(mol\right)\\ PTHH:CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Vì:\dfrac{0,095}{1}< \dfrac{0,1}{1}\Rightarrow H_2dư\\ n_{Cu}=n_{CuO}=0,095\left(mol\right)\\ m_{Cu}=0,095.64=6,08\left(g\right)\)

\(n_{HCl}=\dfrac{18.25}{36.5}=0.5\left(mol\right)\)

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

\(b.\)

\(n_{Mg}=n_{H_2}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0.5=0.25\left(mol\right)\)

\(m_{Mg}=0.25\cdot24=6\left(g\right)\)

\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)

\(c.\)

\(V_{H_2\left(tt\right)}=5.6\cdot90\%=5.04\left(l\right)\)

Fe (0,2 mol) + S \(\underrightarrow{t^o}\) FeS (0,2 mol).

Khối lượng muối FeS thu được là 0,2.88=17,6 (g).

FeS (0,2 mol) + 2HCl → FeCl₂ + H₂S\(\uparrow\) (0,2 mol).

Thể tích khí thu được ở đktc là 0,2.22,4=4,48 (lít).

$a\big)2Al+6HCl\to 2AlCl_3+3H_2$

$b\big)$

$n_{Al}=\dfrac{5,4}{27}=0,2(mol)$

Theo PT: $n_{H_2}=\dfrac{3}{2}n_{Al}=0,3(mol)$

$\to V_{H_2(đktc)}=0,3.22,4=6,72(l)$

$c\big)$

Theo PT: $n_{AlCl_3}=n_{Al}=0,2(mol)$

$\to m_{AlCl_3}=0,2.133,5=26,7(g)$

\(n_{CuO}=\dfrac{16}{80}=0.2\left(mol\right)\)

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

\(0.2........0.2..................0.2\)

\(V_{dd_{H_2SO_4}}=\dfrac{0.2}{1}=0.2\left(l\right)\)

\(m_{CuSO_4}=0.2\cdot160=32\left(g\right)\)

`a)PTHH:`

`Mg + 2HCl -> MgCl_2 + H_2 \uparrow`

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

`b)n_[Mg]=[4,8]/24=0,2(mol)`

  `=>V_[H_2]=0,2.22,4=4,48(l)`

`c)`

`CuO + H_2 -> Cu + H_2 O`

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

   `=>m_[CuO]=0,2.80=16(g)`

\(n_{Fe}=\dfrac{11,2}{56}=0,2(mol)\\ a,Fe+2HCl\to FeCl_2+H_2\\ b,n_{FeCl_2}=n_{H_2}=n_{Fe}=0,2(mol)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48(l)\\ m_{FeCl_2}=0,2.127=25,4(g)\)

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

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

\(0.2....................0.2........0.2\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{FeCl_2}=0.2\cdot127=25.4\left(g\right)\)

\(n_{Fe}=\dfrac{m}{M}=\dfrac{8,4}{56}=0,15mol\)

PTHH: Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

 TL:        1        2           1         1

mol:     0,15 \(\rightarrow\) 0,3 \(\rightarrow\) 0,15 \(\rightarrow\) 0,15

Đổi \(100ml=0,1l\)

\(b.C_{M_{ddHCl}}=\dfrac{n}{V_{dd}}=\dfrac{0,3}{0,1}=3M\)

\(c.V_{H_2}=n.22,4=0,15.22,4=33,6l\)

d. Ta có: \(n_{H_2}=0,15mol\)

PTHH: H₂ + CuO \(\rightarrow\) Cu + H₂O

  TL:     1         1          1        1

mol:  0,15 \(\rightarrow\) 0,15 \(\rightarrow\) 0,15 \(\rightarrow\) 0,15

\(n_{CuO}=\dfrac{m}{M}=\dfrac{20}{80}=0,25mol\)

Lập tỉ lệ: \(\dfrac{n_{H_2}}{1}:\dfrac{n_{CuO}}{1}\)

      \(\Leftrightarrow=\dfrac{0,15}{1}< \dfrac{0,25}{1}\)

\(\Rightarrow\) H₂ hết, CuO dư \(\Rightarrow\) Tính theo H₂

\(m_{CuO}=n.M=0,15.64=9,6g\)