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

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{Cl_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\\ a,V_{Cl_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,m_{FeCl_3}=162,5.0,2=32,5\left(g\right)\)

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

PT: \(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

Theo PT: \(n_{Cl_2}=\dfrac{3}{2}n_{Fe}=0,3\left(mol\right)\)

\(\Rightarrow V_{Cl_2}=0,3.22,4=6,72\left(l\right)\)

b, \(n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)

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

Chọn đáp án A

n F e C l 3 = 32 , 5 162 , 5 = 0 , 2 m o l

Bảo toàn nguyên tố Cl

2 n C l 2 = 3 n F e C l 3 ⇒ n C l 2 = 0 , 3 m o l ⇒ V = 0 , 3.22 , 4 = 6 , 72 l

Chúc em học tốt !!!

\(a.\)

\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)

\(Na+\dfrac{1}{2}Cl_2\underrightarrow{t^0}NaCl\)

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

\(V_{Cl_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{NaCl}=0.2\cdot58.5=11.7\left(g\right)\)

\(b.\)

\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

\(Fe+\dfrac{3}{2}Cl_2\underrightarrow{t^0}FeCl_3\)

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

\(V_{Cl_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_{FeCl_3}=0.1\cdot162.5=16.25\left(g\right)\)

\(c.\)

\(n_{Cu}=\dfrac{6.4}{64}=0.1\left(mol\right)\)

\(Cu+Cl_2\underrightarrow{t^0}CuCl_2\)

\(0.1......0.1.....0.1\)

\(V_{Cl_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{CuCl_2}=0.1\cdot135=13.5\left(g\right)\)

Bài 1:

a. \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

\(2Na+Cl_2\rightarrow2NaCl\)

0,2 ...... 0,1 ..... 0,2 (mol) 

\(\rightarrow\left\{{}\begin{matrix}V_{Cl_2}=0,1.22,4=2,24\left(l\right)\\m_{NaCl}=0,2.58,5=11,7\left(g\right)\end{matrix}\right.\)

b. \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(2Fe+3Cl_2\rightarrow2FeCl_3\)

0,1 ...... 0,15 ...... 0,1 (mol) 

\(\rightarrow\left\{{}\begin{matrix}V_{Cl_2}=0,15.22,4=3,36\left(l\right)\\m_{FeCl_3}=0,1.162,5=16,25\left(g\right)\end{matrix}\right.\)

c. \(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)

\(Cu+Cl_2\rightarrow CuCl_2\)

0,1 .... 0,1 ..... 0,1 (mol) 

\(\rightarrow\left\{{}\begin{matrix}V_{Cl_2}=0,1.22,4=2,24\left(l\right)\\m_{CuCl_2}=0,1.135=13,5\left(g\right)\end{matrix}\right.\)

Đáp án A

Áp dụng định luật bảo toàn khối lượng có:

m c h ấ t   t h a m   g i a   =   m c h ấ t   s ả n   p h ẩ m   =   12 , 7   +   0 , 2   =   12 ,   9   g a m .

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

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

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