2Fe + 3Cl2 \(\rightarrow\)2FeCl3 (1)
FeCl3 + 3AgNO3 \(\rightarrow\)Fe(NO3)3 + 3AgCl (2)
nFe=\(\dfrac{22,4}{56}=0,4\left(mol\right)\)
nAgNO3=\(\dfrac{255}{170}=1,5\left(mol\right)\)
Theo PTHH 1 ta có:
\(\dfrac{3}{2}\)nFe=nCl2=0,6(mol)
nFe=nFeCl3=0,4(mol)
VCl2=0,6.22,4=13,44(lít)
Theo PTHH 2 ta có:
nFeCl3=nFe(NO3)3=0,4(mol)
3nFeCl3=nAgNO3(đã PƯ)=nAgCl=1,2(mol)
nAgNO3 còn dư=1,5-1,2=0,3(mol)
mFe(NO3)3=0,4.242=96,8(g)
mAgNO3=170.0,3= 51(g)
mAgCl=143,5.1,2=172,2(g)
PTHH: 2Fe + 3Cl2 \(\rightarrow\) 2FeCl3 (1)
Theo pt: 2 ........ 3 ........... 2 .... (mol)
Theo đề: 0,4 ... 0,6 ........ 0,4 ... (mol)
PTHH: FeCl3 + 3AgNO3 \(\rightarrow\) Fe(NO3)3 + 3AgCl\(\downarrow\) (2)
Theo pt: .. 1 ............ 3 ............... 1 ............... 3 ...... (mol)
Theo đề: 0,4 .......... 1,2 ............ 0,4 ............ 1,2 ..... (mol)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(V_{Cl_2}=n.22,4=0,6.22,4=13,44\left(l\right)\)
\(n_{AgNO_3}=\dfrac{m}{M}=\dfrac{255}{170}=1,5\left(mol\right)\)
So sánh \(\dfrac{n_{FeCl_{3\left(2\right)đề}}}{n_{FeCl_{3\left(2\right)pt}}}< \dfrac{n_{AgNO_{3_{đề}}}}{n_{AgNO_{3_{pt}}}}\left(\dfrac{0,4}{1}< \dfrac{1,5}{3}\right)\)
=> AgNO3 dư, tính theo nFeCl3
=> Các chất thu được sau phản ứng gồm AgNO3 dư và Fe(NO3)3, AgCl sinh ra.
\(m_{AgNO_{3\left(dư\right)}}=n.M=\left(1,5-1,2\right).170=51\left(g\right)\)
\(m_{Fe\left(NO_3\right)_3}=n.M=0,4.242=96,8\left(g\right)\)
\(m_{AgCl}=n.M=1,2.143,5=172,2\left(g\right)\)
