a) $Fe + 2HCl \to FeCl_2 + H_2$
b) Theo PTHH : $n_{FeCl_2} = n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$m_{FeCl_2} = 0,2.127 = 25,4(gam)$

c) $n_{H_2} = n_{Fe} = 0,2(mol)$
$V_{H_2} = 0,2.24,79 = 4,958(lít)$

d) $RO + H_2 \xrightarrow{t^o} R + H_2O$

Theo PTHH : $n_{RO} = n_{H_2} = 0,2(mol)$
$\Rightarrow M_{RO} = R + 16 = \dfrac{16}{0,2} = 80$
$\Rightarrow R = 64(Cu)$

CTHH oxit : $CuO$
$n_{Cu} = n_{H_2} = 0,2(mol) \Rightarrow m_{Cu} = 0,2.64 = 12,8(gam)$

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

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,2                   0,2      0,2

b) \(V_{H_2}=0,2.24,79=4,958\left(l\right)\)

c) \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

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

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

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

\(M+2HCl\rightarrow MCl_2+H_2\\ n_M=\dfrac{0,486}{M}mol\\ n_{H_2}=\dfrac{0,4958}{24,79}=0,02mol\\ \Rightarrow\dfrac{0,486}{M}=0,02\\ \Rightarrow M\approx24g/mol\\ \Rightarrow M.là.Mg\)

a, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

b, Ta có: \(n_{KMnO_4}=\dfrac{3,16}{158}=0,02\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,01\left(mol\right)\)

\(\Rightarrow m_{O_2}=0,01.32=0,32\left(g\right)\)

c, \(V_{O_2}=0,01.24,79=0,2479\left(l\right)\)

25 độ , 1bar là điều kiện chuẩn

\(PTHH:Zn+H_2SO_4->ZnSO_4+H_2\)

                0,25----------------------------->0,25   (mol)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

\(V_{H_2\left(dkc\right)}=n\cdot24,79=0,25\cdot24,79=6,1975\left(l\right)\\ =>C\)

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

b) \(n_{Mg}=\dfrac{4,8}{24}=0,2mol\) 

\(n_{HCl}=2.n_{Mg}=0,2.2=0,4mol\)

\(\Rightarrow m_{HCl}=n.M=0,4.36,5=14,6g\)

c) \(n_{H_2}=n_{Mg}=0,2mol\) 

Thể tích khí hidro sinh ra (ở đktc): 

\(V_{H_2}=0,2.24,79=4,958l.\)

\(n_{Fe}=\dfrac{2.8}{56}=0.05\left(mol\right)\)

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

\(0.05................................0.05\)

\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)

\(n_{CuO}=\dfrac{6}{80}=0.075\left(mol\right)\)

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

\(1............1\)

\(0.075......0.05\)

Chất khử : H₂ . Chất OXH : CuO 

\(LTL:\dfrac{0.075}{1}>\dfrac{0.05}{1}\Rightarrow CuOdư\)

\(m_{CuO\left(dư\right)}=\left(0.075-0.05\right)\cdot64=1.6\left(g\right)\)