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Theo đề bài ta có : nH2 = 10,08/22,4 = 0,45 (mol)

a) PTHH :

0,45mol->,9mol->0,45mol

b) khối lượng mạt sắt tham gia phản ứng là :

mFe = 0,45.56 = 25,2(g)

c)

nồng độ mol của dd HCl đã dùng là :

CMddHCl = 0,9/0,15 = 6(M)

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{FeCl_2}=n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)

b, \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,15}=4\left(M\right)\)

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

Theo PT: \(n_{NaOH}=2n_{FeCl_2}=0,6\left(mol\right)\)

\(\Rightarrow V_{NaOH}=\dfrac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe,pư}=n_{FeCl_2}=n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ m_{Fe,pư}=0,3.56=16,8g\\ b.n_{HCl}=0,3.2=0,6mol\\ C_{M_{HCl}}=\dfrac{0,6}{0,15}=4M\\ c.2NaOH+FeCl_2\rightarrow Fe\left(OH\right)_2+2NaCl\\ n_{NaOH}=0,3.2=0,6mol\\ V_{ddNaOH}=\dfrac{0,6}{1}=0,6l=600ml\)

nH2=3.36/22.4=0.15 mol

a) PT: Fe + 2HCl ----> FeCl2 + H2

          0.15    0.3                       0.15

b)mFe=0.15*56=8.4g

c)CM_HCl= 0.3*0.05=6 M

Chúc em học tốt!!!

Fe+2HCl->FeCl2+H2

nH2=0.15(mol)

Theo pthh nFe=nH2->nFe=0.15(mol)

mFe phản ứng:0.15*56=8.4(g)

nHCl=2nH2->nHCl=0.3(mol)

CM=0.3:0.05=6 M

Học tốt nheee

Ta có: \(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

a, \(n_{Fe}=n_{H_2}=0,45\left(mol\right)\Rightarrow m_{Fe}=0,45.56=25,2\left(g\right)\)

b, \(n_{HCl}=2n_{H_2}=0,9\left(mol\right)\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,9}{0,15}=6\left(M\right)\)

c, \(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)

\(4Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)

Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_2}=\dfrac{1}{2}n_{FeCl_2}=\dfrac{1}{2}n_{H_2}=0,225\left(mol\right)\)

\(\Rightarrow m_{Fe_2O_3}=0,225.160=36\left(g\right)\)

cảm ơn nha 

Fe + 2HCl → FeCl₂ + H₂↑

\(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

Theo pT: \(n_{Fe}pư=n_{H_2}=0,45\left(mol\right)\)

\(\Rightarrow m_{Fe}pư=0,45\times56=25,2\left(g\right)\)

Theo PT: \(n_{HCl}=2n_{H_2}=2\times0,45=0,9\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,9}{0,15}=6\left(M\right)\)

\(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

Fe + 2HCl → FeCl₂ + H₂

Theo pt: \(n_{Fe}pư=n_{H_2}=0,45\left(mol\right)\)

\(\Rightarrow m_{Fe}pư=0,45\times56=25,2\left(g\right)\)

Theo pt: \(n_{HCl}=2n_{H_2}=2\times0,45=0,9\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,9}{0,15}=6\left(M\right)\)