Đáp án A

n_NaOH = 0,55 mol ; n_H3PO4 = 0,2 mol

=> 2 < n_NaOH / n_H3PO4 < 3

Các phản ứng :

2NaOH + H₃PO₄ -> Na₂HPO₄ + 2H₂O

  2x        x

3NaOH + H₃PO₄ -> Na₃PO₄ + 3H₂O

   3y              y

=> 2x + 3y = 0,55 ; x + y = 0,2

=> x = 0,05 ; y = 0,15 mol

=> Na₂HPO₄ và Na₃PO₄; 7,1g và 24,6g.

$PTHH:NaOH+HCl\to NaCl+H_2O$

$n_{NaOH}=\dfrac{4}{40}=0,1(mol)$

Tỉ lệ: $n_{NaOH}<n_{HCl}\Rightarrow HCl$ dư

Theo PT: $n_{NaCl}=n_{NaOH}=0,1(mol)$

$\Rightarrow m_{muối}=m_{NaCl}=0,1.58,5=5,85(g)$

\(m_{H_2O}=37,6.1=37,6\left(g\right)\\ m_{ddNaOH}=12,4+37,6=50\left(g\right)\\ C\%_{ddNaOH}=\dfrac{12,4}{50}.100=24,8\%\)

Ta có: \(D_{H_2O}=1\left(g/mol\right)\Rightarrow m_{H_2O}=37,6.1=37,6\left(g\right)\)

\(\Rightarrow C\%_{ddNaOH}=\dfrac{12,4.100\%}{37,6}=32,98\%\)

\(n_{H_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\)

Gọi \(n_{NaOH}=a\left(mol\right)\rightarrow n_{NaCl}=1,5a\left(mol\right)\)

PTHH:

2Na + 2H₂O ---> 2NaOH + H₂

a                             a           0,5a

2Na + 2HCl ---> 2NaCl + H₂

1,5a     1,5a              1,5     0,075a

\(\rightarrow0,5a+0,075a=0,025\\ \rightarrow a=0,2\left(mol\right)\)

\(\rightarrow n_{Na}=0,2+0,2.1,5=0,5\left(mol\right)\\ m_{Na}=0,5.23=11,5\left(g\right)\)

\(\rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2.36,5}{100}=7,3\%\\C_{M\left(HCl\right)}=\dfrac{0,2}{\dfrac{100}{1,1}}=0,002M\end{matrix}\right.\)

Cu +2H2SO4----> CuSO4+SO2+2H2O

nSO2=nCu=12,8/64=0,2 mol.

nNaOH=(125.1,28.25)/100=40 gam =>nNaOH=1 mol

nNaOH/nSO2=1/0,2=5 tạo ra muối Natrisunfit Na2SO3

nNa2SO3=nSO2=0,2=> C_MNa2SO3=0,2/0,125=1,6 M.

em lớp 6 chưa học Hoá Thông cảm :))

Uk

\(BTKl:\) \(m_{BaCl2}+m_{Na2SO4}=m_{BaSO4}+m_{NaCl}\)

\(\Rightarrow m_{NaCl}=m_{BaCl2}+m_{Na2SO4}-m_{NaCl}=20,8+14,2-23,3=11,7\left(g\right)\)

a) \(m_{NaOH}=6\cdot157.5\cdot\dfrac{1.06}{100}=10.017\left(g\right)\)

\(n_{NaOH}=\dfrac{10.017}{40}\simeq0.25\left(mol\right)\)

\(NaOH+CO_2\underrightarrow{t^0}NaHCO_3\)

\(n_{CO_2}=n_{NaOH}=0.25\left(mol\right)\)

\(n_{C_nH_{2n}}=\dfrac{2.8}{22.4}=0.125\left(mol\right)\)

\(C_nH_{2n}+\dfrac{3n}{2}O_2\underrightarrow{t^0}nCO_2+nH_2O\)

\(1.............................n\)

\(0.125..........................0.25\)

\(\Rightarrow n=\dfrac{0.25}{0.125}=2\)

\(CT:C_2H_4\)

\(CH_2=CH_2\)

b) 

\(C_nH_{2n}+Br_2\rightarrow C_nH_{2n}Br_2\)

\(14n..............14n+160\)

\(10.08\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot48.48\)

\(\Rightarrow48.48\cdot14n=10.08\left(14n+160\right)\)

\(\Rightarrow n=3\)

\(CT:C_3H_6\)

\(CH_2=CH-CH_3\)

Chúc bạn học tốt !!

a, \(n_P=\dfrac{7,75}{31}=0,25\left(mol\right)\)

PTHH: 4P + 5O₂ ---t^o→ 2P₂O₅

Mol:    0,25                      0,125      

PTHH: P₂O₅ + 3H₂O → 2H₃PO₄

Mol:     0,125                      0,25

\(C\%_{ddH_3PO_4}=\dfrac{0,25.98.100\%}{280}=8,75\%\)

b, \(n_{NaOH}=0,3.2=0,6\left(mol\right)\)

Ta có: \(T=\dfrac{0,6}{0,25}=2,4\) ⇒ tạo ra 2 muối Na₂HPO₄, Na₃PO₄

PTHH: 2NaOH + H₃PO₄ → Na₂HPO₄ + 2H₂O

Mol:         2x           x                  x

PTHH: 3NaOH + H₃PO₄ → Na₃PO₄ + 3H₂O

Mol:         3y            y                 y

Ta có: \(\left\{{}\begin{matrix}x+y=0,25\\2x+3y=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)

\(m_{muối}=0,15.141+0,1.164=37,55\left(g\right)\)