Phương trình hoá học : 2Fe + 3 Cl 2 → t ° FeCl 3

Theo định luật bảo toàn khối lượng :

m Fe + m Cl 2 = m FeCl 3

  m Cl 2 = m FeCl 3 - m Fe = 16,25 - 5,6 = 10,65g

a) \(n_{Cl_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 2Fe + 3Cl₂ --t^o--> 2FeCl₃

_____\(\dfrac{2}{15}\)<--0,2------------->\(\dfrac{2}{15}\)

=> m_Fe = \(\dfrac{2}{15}.56=7,467\left(g\right)\)

b) \(m_{FeCl_3}=\dfrac{2}{15}.162,5=21,667\left(g\right)\)

a+b) PTHH: \(2Fe+3Cl_2\xrightarrow[]{t^o}2FeCl_3\)

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

Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cl_2}=0,3\left(mol\right)\\n_{HCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Cl_2}=0,3\cdot71=21,3\left(g\right)\\V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)=200\left(ml\right)\end{matrix}\right.\) 

Câu 1:

\(2Na+Br_2\rightarrow2NaBr\\ n_{NaBr}=\dfrac{61,8}{103}=0,6\left(mol\right)\\ n_{Na}=n_{NaBr}=0,6\left(mol\right)\\ n_{Br_2}=\dfrac{0,6}{2}=0,3\left(mol\right)\\ \Rightarrow a=m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Br_2}=0,3.160=48\left(g\right)\\ m_{ddBr_2}=\dfrac{48}{5\%}=960\left(g\right)\)

Câu 2:

\(2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{FeCl_3}=\dfrac{40,625}{162,5}=0,25\left(mol\right)\\ n_{Fe}=n_{FeCl_3}=0,25\left(mol\right)\\ \Rightarrow m=m_{Fe}=0,25.56=14\left(g\right)\\ n_{Cl_2}=\dfrac{3}{2}.0,25=0,375\left(mol\right)\\ V_{Cl_2\left(đktc\right)}=0,375.22,4=8,4\left(l\right)\)

\(n_{C_6H_5Cl}=\dfrac{78}{112,5}=\dfrac{52}{75}\left(mol\right)\)

\(n_{C_6H_6}=\dfrac{78}{78}=1\left(mol\right)\)

PTHH:

\(C_6H_6+Cl_2\xrightarrow[t^o]{Fe}C_6H_5Cl\)

1------------------>1

\(\Rightarrow H=\dfrac{\dfrac{52}{75}}{1}.100\%=69,33\%\)

Ta có: \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)

a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

______0,5____0,5_____0,5_____0,5 (mol)

b, m_H2SO4 = 0,5.98 = 49 (g)

c, m_FeSO4 = 0,5.152 = 76 (g)

d, \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

____0,5__0,5 (mol)

⇒ m_Cu = 0,5.64 = 32 (g)

Bạn tham khảo nhé!

$2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$
$n_{FeCl_3} = \dfrac{32,5}{162,5} = 0,2(mol)$

$n_{Cl_2} = \dfrac{3}{2}n_{FeCl_3} = 0,3(mol)$
$V_{Cl_2} = 0,3.22,4 = 6,72(lít)$

\(n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)

        1         2             1           1

     0,05                                 0,05

\(n_{Fe}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

⇒ \(m_{Fe}=0,05.56=2,8\left(g\right)\)

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