Đổi 200ml = 0,2 lít

Ta có: \(C_{M_{H_2SO_4}}=\dfrac{n_{H_2SO_4}}{0,2}=1,5M\)

=> \(n_{H_2SO_4}=0,3\left(mol\right)\)

a. PTHH: Al₂O₃ + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂O

Theo PT: \(n_{Al_2O_3}=\dfrac{1}{3}.n_{H_2SO_4}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)

=> \(a=m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

b. Theo PT: \(n_{Al_2\left(SO_4\right)_3}=n_{Al}=0,1\left(mol\right)\)

=> \(m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)

Bài 13 : 

\(a)n_{Fe_2O_3} = \dfrac{9,6}{160} = 0,06(mol)\\ Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O\\ n_{HCl} = 6n_{Fe_2O_3} = 0,36(mol)\\ C\%_{HCl} = \dfrac{0,36.36,5}{150}.100\% = 8,76\%\\ \Rightarrow X = 8,76 b) n_{FeCl_3} = 2n_{Fe_2O_3} = 0,12(mol)\\ m_{FeCl_3} = 0,12.162,5 =19,5(gam)\)

Bài 11 : 

\(a) n_{NaOH} = 0,2.1,5 = 0,3(mol)\\ 2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O\\ n_{H_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,15(mol)\\ \Rightarrow x = \dfrac{0,15}{0,12}= 1,25(M)\\ b) n_{Na_2SO_4} = n_{H_2SO_4} = 0,15(mol)\\ m_{Na_2SO_4} = 0,15.142 = 21,3(gam)\)

Bài 12 : 

\(a)n_{Al} = \dfrac{16,2}{27} = 0,6(mol)\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2SO_4} = n_{H_2} = \dfrac{3}{2}n_{Al} = 0,9(mol)\\ \Rightarrow X = \dfrac{0,9}{0,2} =4,5(M)\\ b) V_{H_2} = 0,9.22,4 = 20,16(lít)\)

Chọn đáp án A

\(n_{HCl}=C_M.V=1,5.0,2=0,3\left(mol\right)\)

PTHH :

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

  0,1       0,3           0,1

\(C_{M_{AlCl_3}}=\dfrac{n}{V}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

\(a,2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{HCl}=0,2.1,5=0,6\left(mol\right)\\ n_{H_2}=\dfrac{3}{6}.0,6=0,3\left(mol\right);n_{Al}=\dfrac{2}{6}.0,6=0,2\left(mol\right)\\ b,V=V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ m=m_{Al}=0,2.27=5,4\left(g\right)\)

Chọn B

Đáp án là B. 11,2 gam

Đáp án : B

Fe + 2HCl  à FeCl₂ + H₂

=> n_Fe = n_FeCl2 = 0,2 mol

=> m = 11,2g