\(Fe+H_2SO_4 \to FeSO_4+H_2\\ n_{H_2}=0,15(mol)\\ a/\\ n_{Fe}=n_{H_2}=0,15(mol)\\ m_{Fe}=0,15.56=8,4(g)\\ b/\\ n_{H_2SO_4}=n_{H_2}=0,15(mol)\\ CM_{H_2SO_4}=\dfrac{0,15}{2}=0,75M c/\\ n_{FeSO_4}=n_{H_2}=0,15(mol)\\ CM_{FeSO_4}=\dfrac{0,15}{0,2}=0,75M\\\)

a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

____0,15<--0,3<-------------0,15

=> m_Fe = 0,15.56 = 8,4 (g)

b) \(C_{M\left(ddHCl\right)}=\dfrac{0,3}{0,05}=6M\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,,15(mol)\\ Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow n_{Fe}=n_{H_2SO_4}=0,15(mol)\\ a,m_{Fe}=0,15.56=8,4(g)\\ b,C_{M_{H_2SO_4}}=\dfrac{0,15}{0,8}=0,1875M\)

Câu 3 : 

\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)

        1          1                1           1

      0,15       0,15                       0,15

a) \(n_{Fe}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

⇒ \(m_{Fe}=0,15.56=8,4\left(g\right)\)

b) \(n_{H2SO4}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

800ml = 0,8l

\(C_{M_{ddH2SO4}}=\dfrac{0,15}{0,8}=0,1875\left(M\right)\)

 Chúc bạn học tốt

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

a) \(n_{Fe}=n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ \Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

b) \(n_{H_2SO_4}=n_{HCl}=0,15\left(mol\right)\\ \Rightarrow CM_{H_2SO_4}=\dfrac{0,15}{0,8}=0,1875M\)

a)

$Fe + 2HCl \to FeCl_2 + H_2$

b) Theo PTHH : $n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$m_{Fe} = 0,15.56 = 8,4(gam)$

c) $n_{HCl} = 2n_{H_2} = 0,3(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,3}{0,05} = 6M$

d) $2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$

$n_{H_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,25(mol)$

$m_{dd\ H_2SO_4} = \dfrac{0,25.98}{20\%} = 122,5(gam)$

$V_{dd\ H_2SO_4} = \dfrac{122,5}{1,14} = 107,5(ml)$

\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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

            1         2             1          1

          0,15     0,3                      0,15

b) \(n_{Fe}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

⇒ \(m_{Fe}=0,15.56=8,4\left(g\right)\)

c) \(n_{HCl}=\dfrac{0,15.2}{1}=0,3\left(mol\right)\)

50ml = 0,05l

\(C_{M_{ddHCl}}=\dfrac{0,3}{0,05}=6\left(M\right)\)

 Chúc bạn học tốt

Chọn B

Fe+H2SO4→FeSO4+H2

nH2=0,15(mol)

a/

nFe=nH2=0,15(mol)

mFe=0,15.56=8,4(g)

b/

 )

800ml = 0,8l