\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)

tỉ lệ:          1      :    2     :     1        :     1

n(mol)      0,2--->0,4------->0,2----->0,2

\(m_{FeCl_2}=n\cdot M=0,2\cdot\left(56+35,5\cdot2\right)=25,4\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\\ V_{H_2\left(dkc\right)}=n\cdot24,79=4,958\left(l\right)\)

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

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

Theo phương trình hóa học: \(n_{H_2}=n_{Fe}=o,2\left(mol\right)\)

\(V_{H_2\left(dktc\right)}=n_{H_2}\times22,4=0,2\times22,4=4,48\left(l\right)\)

\(V_{H_2\left(dkc\right)}=n_{H_2}\times24,79=0,2\times24,79=4,96\left(l\right)\)

c) Theo phương trình hóa học: \(n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\)

\(m_{FeCl_2}=n_{FeCl_2}\times M_{FeCl_2}=0,2.127=25,4\left(g\right)\)

a, \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Theo PT: \(n_{H_2}=n_{Mg}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, \(n_{HCl}=2n_{Mg}=0,2\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,2}{2}=0,1\left(l\right)\)

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PTHH :

\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

0,1        0,2                        0,1 

\(a,V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(b,V=\dfrac{n}{C_M}=\dfrac{0,2}{2}=0,1\left(l\right)\)

BTKL: \(m_{Fe}+m_{HCl}=m_{muối}+m_{H_2}\)

 \(\Rightarrow m_{H_2}=5,6+7,3-12,7=0,2\left(g\right)\)

Gọi \(n_{Fe}=x\left(mol\right)\)

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

\(n_{FeCl_2}=n_{Fe}=x\left(mol\right)\)

Vì khối lượng muối FeCl2 tăng 7,1g so với khối lượng bột Fe

\(\Rightarrow127x-56x=7,1\\ \Rightarrow x=0,1\)

\(n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ V_{H_2\left(ĐKTC\right)}=0,1.22,4=2,24\left(l\right)\)

Chọn D

a. \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Theo PTHH: \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\)

Khối lượng muối thu được:

\(m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

b. Theo PTHH: \(n_{H_2}=\dfrac{n_{Al}.3}{2}=\dfrac{0,2.3}{2}=0,3\left(mol\right)\)

Thể tích khí hiđro thu được:

\(V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72\left(l\right)\)

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

b) nFe = \(\dfrac{11,2}{56}=0,2mol\)

Theo pt: nH₂ = nFe = 0,2 mol

=> VH₂ = 0,2.22,4 = 4,48lit

c) Theo pt: nHCl = 2nFe = 0,4 mol

=> mHCl = 0,4.36,5 = 14,6 g

=> C% = \(\dfrac{14,6}{73}.100\%=20\%\)

a) Pt: $Fe+2HCl\to FeC{}^{}+{}^{}$

b) nFe = \(\dfrac{11,2}{56}=0,2mol\)

Theo pt: nH₂ = nFe = 0,2 mol

=> VH₂ = 0,2.22,4 = 4,48lit

c) Theo pt: nHCl = 2nFe = 0,4 mol

=> mHCl = 0,4.36,5 = 14,6 g

=> \(C\%=\dfrac{14,6}{73}.100\%=20\%\)

a) nFe=0,1(mol); nHCl=0,4(mol)

PTHH: Fe + 2 HCl -> FeCl2 + H2

Ta có: 0,1/1 < 0,4/2

=> Fe hết, HCl dư, tish theo nFe.

b) nH2=nFeCl2=Fe=0,1(mol)

=> V(H2,đktc)=0,1.22,4=2,24(l)

c) mFeCl2=127.0,1=12,7(g)

a) nFe=0,1(mol); nHCl=0,4(mol) PTHH: Fe + 2 HCl -> FeCl2 + H2 Ta có: 0,1/1 < 0,4/2 => Fe hết, HCl dư, tish theo nFe. b) nH2=nFeCl2=Fe=0,1(mol) => V(H2,đktc)=0,1.22,4=2,24(l) c) mFeCl2=127.0,1=12,7(g)