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

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

\(n_{Fe}=\dfrac{6.72}{56}=0.12\left(mol\right)\)

\(n_{Cl_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

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

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

\(m_{Fe\left(dư\right)}=\left(0.12-0.1\right)\cdot56=1.12\left(g\right)\)

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

a) PTHH: FeCl₃ + 3KOH → Fe(OH)₃ + 3KCl

b) Theo ĐLBTKL ta có:

 \(m_{FeCl_3}+m_{KOH}=m_{Fe\left(OH\right)_3}+m_{KCl}\)

\(\Leftrightarrow m_{FeCl_3}=m_{Fe\left(OH\right)_3}+m_{KCl}-m_{KOH}=2,14+4,47-3,36=3,25\left(g\right)\)

\(a.Sắt+Clo\rightarrow Sắt\left(III\right)clorua\\ b.2Fe+3Cl_2-^{t^o}\rightarrow2FeCl_3\\ c.m_{Fe}+m_{Cl_2}=m_{FeCl_3}\\ \Rightarrow m_{FeCl_3}=5,6+10,65=16,25\left(g\right)\)

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

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

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

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

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

a.3.36l

b.13.25g

câu b là 16,25g nha bạn oi mik ghi nhầm

\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\\ pthh:Mg+H_2SO_4->MgSO_4+H_2\)
          0,25    0,25           0,25            0,25 
\(m_{MgSO_4}=0,25.120=30\left(g\right)\)
\(n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\\ pthh:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
  LTL : \(\dfrac{0,15}{1}>\dfrac{0,25}{3}\)
=> Fe dư , H2 hết 
=> \(m_{Fe}=\dfrac{1}{6}.56=\approx9,3\left(g\right)\)

a) PTHH là : 2Fe + 3Cl2 → 2FeCl3

b) số mol Fe là:

nFe = 2,8\56=0,05(mol)

pthh : 2Fe + 3Cl2 → 2FeCl3

0,05mol→0,075mol→0,05mol

thể tích Cl2 là :

VCl2 = 0,075. 22,4 = 1,68 l

c) khối lượng FeCl3 là :

mFeCl3 = 0,05 .162,5,= 8,125 g

Bảo toàn KL: \(m_{FeCl_3}+m_{KOH}=m_{Fe\left(OH\right)_3}+m_{KCl}\)

\(\Rightarrow m_{FeCl_3}=7+8,25-5,5=9,75\left(g\right)\)

Chọn D

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

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

        1          2            1           1

       0,2       0,4         0,2        0,2

a) \(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

b) \(n_{FeCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{FeCl2}=0,2.127=25,4\left(g\right)\)

c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

⇒ \(m_{HCl}=0,.4.36,5=14,6\left(g\right)\)

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