NaOH + HCl --> NaCl+H2O (1)

2NaOH +H2SO4 --> Na2SO4 +2H2O (2)

Đặt n_{NaOH (1)}= a(mol)

n_NaOH(2) = b (mol)

=>a + b = 0,3.2 =0,6( *)

Theo PT (1) : n_NaCl = n_NaOH(1) = a(mol)

Theo PT (2) : n_Na2SO4=\(\dfrac{1}{2}\) n_NaOH(2) = 0,5b(mol)

=>58,5a + 71b =40,1(**)

Từ (*), (**) => a= 0,2 ; b = 0,4 

 n_HCl = n_{NaOH (1)}=0,2 mol 

=> \(x=CM_{HCl}=\dfrac{0,2}{0,2}=1M\)

n_H2SO4 = n_{NaOH (2)}=0,4 mol 

\(y=CM_{H_2SO_4}=\dfrac{0,4}{0,2}=2M\)

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 : 

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

Ta có: \(n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,2.1+2.0,2.0,5=0,4\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

⇒ n_OH^- = n_H⁺ = 0,4 (mol)

Mà: \(n_{OH^-}=n_{NaOH}+2n_{Ba\left(OH\right)_2}=0,3.0,5+2.0,3.a\left(mol\right)\)

⇒ 0,3.0,5 + 2.0,3.a = 0,4 ⇒ a = 5/12 (M)

\(n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,3.\dfrac{5}{12}=0,125\left(mol\right)\)

\(n_{SO_4^{2-}}=n_{H_2SO_4}=0,2.0,5=0,1\left(mol\right)\)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)

\(\Rightarrow n_{BaSO_4}=0,1\left(mol\right)\Rightarrow m=m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

Ta có: \(n_{NaOH}=0,3\cdot2=0,6\left(mol\right)\)

Bảo toàn Clo và Natri: \(n_{NaOH}=n_{NaCl}=n_{HCl}=0,6\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{0,6}{0,2}=3\left(M\right)\\m_{NaCl}=0,6\cdot58,5=35,1\left(g\right)\end{matrix}\right.\)

trong dung dịch ban đầu nhé mn

NaOH + HCl --> NaCl+H2O (1)

2NaOH +H2SO4 --> Na2SO4 +2H2O (2)

Đặt n_{NaOH (1)}= a(mol)

n_NaOH(2) = b (mol)

=>a + b = 0,3.2 =0,6( *)

Theo PT (1) : n_NaCl = n_NaOH(1) = a(mol)

Theo PT (2) : n_Na2SO4=12

 n_NaOH(2) = 0,5b(mol)

=>58,5a + 71b =40,1(**)

Từ (*), (**) => a= 0,2 ; b = 0,4 

 n_HCl = n_{NaOH (1)}=0,2 mol 

=> x=CMHCl=0,20,2=1M

n_H2SO4 = n_{NaOH (2)}=0,4 mol 

y=CMH2SO4=0,40,2=2M

Chọn B