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

\(n_{Fe_2O_3}=\dfrac{16}{160}=0.1\left(mol\right)\)

\(2Fe+\dfrac{3}{2}O_2\underrightarrow{t^0}Fe_2O_3\)

\(0.2....0.15.........0.1\)

\(n_{Fe\left(pư\right)}=0.2\left(mol\right)< 0.3\Rightarrow Fedư\)

\(V_{O_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_{Fe\left(dư\right)}=\left(0.3-0.2\right)\cdot56=5.6\left(g\right)\)

                     3Fe + 2O₂ → Fe₃O₄

_{Theo pt   :          3            2              1          mol}

_{Theo đề bài :       0,2         0,3        0,2/3}

_a.

Ta có tỉ lệ \(\dfrac{0,2}{3}< \dfrac{0,3}{2}\) _{nên  Fe phản ứng hết , oxi dư số mol sắt từ thu được tính theo Fe}

b. nFe₃O₄ = 0,2/3 mol ==> m Fe₃O₄ = 0,2 /3 .232 = 15,47 gam

nHCl=2.0,1=0,2mol

PTHH: Fe+2HCl=>FeCl2+H2

           0,1<-0,2--->0,1---->0,1

mFe phản ứng=0,1.56=5,6g

VH2 thu đc=0,1.22,4=2,24 lít

a) n Fe = 16,8/56 = 0,3(mol)

$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$

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

=> V H2 = 0,45.22,4 = 10,08(lít)

b)

$Fe + 2HCl \to FeCl_2 + H_2$

n HCl = 2n Fe = 0,6(mol)

=> m HCl = 0,6.36,5 = 21,9 gam

a, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

Ta có: \(n_{KClO_3}=\dfrac{24,5}{122,5}=0,2\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,3\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,3.24,79=7,437\left(g\right)\)

b, PT: \(2Cu+O_2\underrightarrow{t^o}2CuO\)

Ta có: \(n_{Cu}=\dfrac{32}{64}=0,5\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,5}{2}< \dfrac{0,3}{1}\), ta được O₂ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{Cu}=0,25\left(mol\right)\\n_{CuO}=n_{Cu}=0,5\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,3-0,25=0,05\left(mol\right)\)

\(\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)

\(m_{CuO}=0,5.80=40\left(g\right)\)

`a)PTHH:`

`Fe + 2HCl -> FeCl_2 + H_2`

`0,3`    `0,6`          `0,3`     `0,3`       `(mol)`

`n_[Fe]=[22,4]/56=0,4(mol)`

`n_[HCl]=0,3.2=0,6(mol)`

Ta có:`[0,4]/1 > [0,6]/2`

   `=>Fe` dư

`b)m_[FeCl_2]=0,3.127=38,1(g)`

`c)m_[Fe(dư)]=(0,4-0,3).56=5,6(g)`

\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ n_{HCl}=0,3.2=0,6\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\\ LTL:\dfrac{0,4}{1}>\dfrac{0,6}{2}\) 
=> Fe dư 
\(n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\\ m_{FeCl_2}=0,3.127=38,1\left(g\right)\\ n_{Fe\left(p\text{ư}\right)}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\\ m_{Fe\left(d\right)}=\left(0,4-0,3\right).56=5,6\left(g\right)\)

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

\(n_{HCl}=0,3.2=0,6\left(mol\right)\)

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

Xét: \(\dfrac{0,4}{1}>\dfrac{0,6}{2}\)                            ( mol )

         0,3      0,6          0,3               ( mol )

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

\(m_{Fe\left(dư\right)}=\left(0,4-0,3\right).56=5,6\left(g\right)\)

PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{3}< \dfrac{0,4}{2}\) \(\Rightarrow\) Oxi còn dư, Fe p/ứ hết

\(\Rightarrow n_{O_2\left(dư\right)}=0,4-\dfrac{2}{15}=\dfrac{4}{15}\left(mol\right)\)

+) Theo PTHH: \(\left\{{}\begin{matrix}n_{O_2}=\dfrac{2}{15}\left(mol\right)\\n_{Fe_3O_4}=\dfrac{1}{15}\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{kk}=\dfrac{2}{15}\cdot22,4\cdot5\approx14,93\left(l\right)\\m_{Fe_3O_4}=\dfrac{1}{15}\cdot232\approx15,47\left(g\right)\end{matrix}\right.\)