\(a) Fe + 2HCl \to FeCl_2 + H_2\\ n_{FeCl_2} = n_{H_2} = n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ b)m_{FeCl_2} = 0,1.127 = 12,7(gam)\\ c) n_{HCl} =2 n_{Fe} = 0,2(mol)\\ C_{M_{HCl}} = \dfrac{0,2}{0,2} = 1M\)

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

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

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

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{FeCl_2}=0.1\cdot127=12.7\left(g\right)\)

\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)

\(a) Fe + 2HCl \to FeCl_2 + H_2\\ b) n_{H_2} = n_{Fe} = \dfrac{8,4}{56} = 0,15(mol)\\ V_{H_2} = 0,15.22,4 = 3,36(lít)\\ n_{HCl} = 2n_{H_2} = 0,3(mol)\ m_{HCl} = 0,3.36,5 = 10,95(gam)\)

\(n_{Fe}=\dfrac{8.4}{56}=0.15\left(mol\right)\)

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

\(0.15......0.3..................0.15\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

Fe + 2HCl -> FeCl2 + H2
nFe = 5,6/56 = 0,1 mol
=>nH2 = 0,1 mol
=> VH2= 0,1*22,4= 2,24 lít

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

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

          0,1-->0,2------------------>0,1

=> \(\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,2.36,5}{15\%}=\dfrac{146}{3}\left(g\right)\\V_{H_2}=0,1.22,4=4,48\left(l\right)\end{matrix}\right.\)

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

            \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

a+b) Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)=n_{H_2SO_4}=n_{ZnSO_4}\) 

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,3}{0,2}=1,5\left(M\right)=C_{M_{ZnSO_4}}\)

c) Theo PTHH: \(n_{H_2}=n_{Zn}=0,3mol\) \(\Rightarrow V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\)

d) Theo PTHH: \(n_{Fe}=\dfrac{2}{3}n_{H_2}=0,2mol\)

\(\Rightarrow m_{Fe}=0,2\cdot56=11,2\left(g\right)\)

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Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

____0,1______0,2_____0,1____0,1 (mol)

a, \(C_{M_{HCl}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}\left(M\right)\)

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

b, \(FeCl_2+2NaOH\rightarrow2NaCl+Fe\left(OH\right)_2\)

Theo PT: \(n_{NaOH}=2n_{FeCl_2}=0,2\left(mol\right)\)

\(\Rightarrow V_{NaOH}=\dfrac{0,2}{2}=0,1\left(l\right)\)

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)

\(a) Fe + 2HCl \to FeCl_2 + H_2\\ b) n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)\\ n_{HCl} = 2n_{Fe} = 0,1.2 = 0,2(mol)\\ m_{HCl} = 0,2.36,5 = 7,3(gam)\\ m_{dung\ dịch\ HCl} = \dfrac{7,3}{10\%} = 73(gam)\)