Bài 32: Hiđro sunfua-Lưu huỳnh Đioxit - Lưu huỳnh Trioxit

ta có : \(n_{Na_2SO_3}=n_{SO_2}=0,05\)

\(\Rightarrow V_{SO_2}=0,05.22,4=1,12\left(l\right)\)

n\(_{SO_2}\)=\(\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

n\(_{NaOH}\)=0,8.0,25=0,2(mol)

T=\(\dfrac{0,2}{0,25}=0,8< 1\).vậy phản ứng tạo ra 1 muối là NaHS\(O_3\) và dư SO\(_2\):

\(SO_2+NaOH\rightarrow NaHSO_3\)

bđ: 0,25 0,2 (mol)

pư:0,2 0,2 0,2 (mol)

dư: 0,05 0 0 (mol)

SO\(_2+H_2O\rightarrow H_2SO_3\)

0,05 0,05 (mol)

\(C_{M_{NaHSO_3}}=\dfrac{0,2}{0,25}=0,8M\)

\(C_{M_{H_2SO_3}}=\dfrac{0,05}{0,25}=0,2M\)

A

A

Natri hiroxit: NaOH

Sắt (II) oxit: FeO

Canxi đihiđrophotphat: Ca(H₂PO₄)₂

NaOH

FeO

Ca₂(H₂PO₄)₃

n_SO2 = 0,2 mol

n_NaOH = 0,3 mol

Ta có

\(\dfrac{n_{NaOH}}{n_{SO2}}\) = \(\dfrac{0,3}{0,2}\) = 1,5

\(\Rightarrow\) tạo 2 muối

2NaOH + SO₂ \(\rightarrow\)Na₂SO₃ + H₂O

x................0,5x.......0,5x

NaOH + SO₂ \(\rightarrow\) NaHSO₃

y..............y.............y

Ta có hệ

\(\left\{{}\begin{matrix}x+y=0,3\\0,5x+y=0,2\end{matrix}\right.\)

\(\Rightarrow\) \(\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

\(\Rightarrow\) m_Na2SO3 = 0,2.126 = 25,2 (g)

\(\Rightarrow\) m_NaHSO3 = 0,1.104 = 10,4 (g)

2Al + 6HCl -> 2AlCl₃ + 3H₂ (1)

Fe + 2HCl -> FeCl₂ -> H₂ (2)

n_H2=0,135(mol)

m_Al,Fe=6-1,86=5,865(g)

Đặt n_Al=a(mol)

n_Fe=b(mol)

Ta có hệ:

\(\left\{{}\begin{matrix}27a+56b=5,865\\1,5a+b=0,135\end{matrix}\right.\)

=>a=0,03;b=0,09

m_Al=0,03.27=0,81(g)

Tự tính %m

b;

ta có:

n_HCl=2n_H2=0,27(mol)

V dd HCl=\(\dfrac{0,27}{2}=0,135\left(lt\right)\)

\(2Al\left(a\right)+6HCl\left(3a\right)\rightarrow2AlCl_3+3H_2\left(1,5a\right)\)

\(Fe\left(b\right)+2HCl\left(2b\right)\rightarrow FeCl_2+H_2\left(b\right)\)

\(m_{Al,Fe}=6-1,86=4,14\left(g\right)\Rightarrow27a+56b=4,14\left(II\right)\)

\(n_{H_2}=0,135\left(mol\right)\Rightarrow1,5a+b=0,135\left(I\right)\)

Từ (I,II) => a=0,06; b = 0,045

\(\Rightarrow\%m\)

\(V_{ddHHCl}=\dfrac{3.0,06+2.0,045}{2}=0,135M\)

Đặt n_NaOH=a

n_KOH=b

Ta có hệ:

\(\left\{{}\begin{matrix}40a+56b=3,04\\58,5a+74,5b=4,15\end{matrix}\right.\)

=>a=0,02;b=0,04

tự tính %

b;

n_{HCl tham gia}=n_OH=0,06(mol)

C_M dd HCl=\(\dfrac{0,06}{0,3}=0,2M\)

Đặt n_NaOH = x (mol); n_KOH = y (mol)

NaOH + HCl \(\rightarrow\) NaCl + H₂O (1)

KOH + HCl \(\rightarrow\) KCl + H₂O (2)

Từ (1)(2) ta có hệ pt

\(\left\{{}\begin{matrix}40x+56y=3,04\\58,5x+74,5y=4,15\end{matrix}\right.\)

\(\Rightarrow\) \(\left\{{}\begin{matrix}x=0,02\\y=0,04\end{matrix}\right.\)

\(\Rightarrow\) %_NaCl = \(\dfrac{0,02.58,5.100}{4,15}\) \(\approx\) 28,2%

\(\Rightarrow\) %_KCl = \(\dfrac{0,04.74,5.100}{4,15}\) \(\approx\) 71,8%

\(\Rightarrow\) C_{M HCl} = \(\dfrac{0,02+0,04}{0,3}\) = 0,2 (M)

n_H2=0,8(mol)

Đặt n_Mg=a

n_Zn=b

n_Fe=2a

Ta có hệ:

\(\left\{{}\begin{matrix}112a+24a+65b=46,1\\3a+b=0,8\end{matrix}\right.\)

=>a=0,1;b=0,5

tự tính % khối lượng

b;

Ta có:

n_HCl=2n_H2=1,6(mol)

V=1,6:2=0,8(lít)