\(4.\)

\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

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

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

\(m_{Fe}=0.15\cdot56=8.4\left(g\right)\)

\(C_{M_{HCl}}=\dfrac{0.3}{0.5}=0.6\left(M\right)\)

\(5.\)

\(Đặt:n_{Fe}=a\left(mol\right),n_{Al}=b\left(mol\right)\)

\(m_{hh}=56a+27b=8.3\left(g\right)\left(1\right)\)

\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

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

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

\(\Rightarrow a+1.5b=0.25\left(2\right)\)

\(\left(1\right),\left(2\right):a=b=0.1\)

\(\%Fe=\dfrac{5.6}{8.3}\cdot100\%=67.47\%\)

\(\%Al=32.53\%\)

Theo gt ta có: $n_{HCl}=0,1(mol)$

$Fe+2HCl\rightarrow FeCl_2+H_2$

a, Ta có: $n_{Fe}=0,05(mol)\Rightarrow m_{Fe}=2,8(g)$

b, Ta có: $n_{H_2}=0,05(mol)\Rightarrow V_{H_2}=1,12(l)$

\(n_{HCl}=C_M.V=0,1mol\)

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

- Theo PTHH : nFe = 0,05mol

=> m = 2,8g

b, - Theo PTHH : nH2 = 0,05mol

=> V = 1,12l
 

Đáp án là D. 6,05

a)

n HCl = 0,5.0,2 = 0,1(mol)

$Fe + 2HCl \to FeCl_2 + H_2$

Theo PTHH : 

n Fe = 1/2 n HCl = 0,05(mol)

=> m = 0,05.56 = 2,8(gam)

b) Theo PTHH : 

n H2 = n Fe = 0,05(mol)

=> V H2 = 0,05.22,4 = 1,12(lít)

Chọn B.

Đáp án A

Hướng dẫn

\(n_{H_2SO_4}=0.6\left(mol\right)\)

\(4Fe^{\dfrac{+3}{4}}\rightarrow4Fe^{3+}+9e\)

\(x...................\dfrac{9}{4}x\)

\(S^{+6}+2e\rightarrow S^{+4}\)

\(0.6......1.2\)

Bảo toàn e : 

\(\dfrac{9}{4}x=1.2\Rightarrow x=\dfrac{8}{15}\)

\(m=\dfrac{8}{15}\cdot232=123.7\left(g\right)\)