a, \(n_{KOH}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)

b, \(n_{KOH\left(trong300mlA\right)}=0,3.0,3=0,09\left(mol\right)\)

Gọi: V_H2O = a (l)

\(\Rightarrow C_{M_A}=0,2=\dfrac{0,09}{a+0,3}\Rightarrow a=0,15\left(l\right)=150\left(ml\right)\)

CTPT oleum: H₂SO₄.nSO₃ có x (mol)

H₂SO₄.nSO₃ + nH₂O → (n + 1)H₂SO₄

x          →                             x(n + 1)

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

x(n + 1)→ 2x(n + 1)

=> (98 + 80n)x = 3,38 và 2x (n + 1) = 0,08

=> x = 0,01 và nx = 0,03

=> n = 3

Suy ra CTPT oleum: H₂SO₄.3SO₃

\(a.\\ C\%_{sau}=\dfrac{5}{100}=\dfrac{32.0,1}{32+m_{H_2O}}\\ m_{H_2O}=32\left(g\right)\\ b.\\ C_{M\left(sau\right)}=1=\dfrac{0,2.2}{0,2+V_{H_2O}}\\ V_{H_2O}=0,2\left(L\right)=200\left(mL\right)\)

m FeSO4=\(\dfrac{27,8.152}{278}=15,2g\)

=>m dd=\(\dfrac{15,2}{\left(9\%\right)}=\dfrac{1520}{9}\)

m H2O=\(\dfrac{1520}{9}\)-27,8=141g

->Vh2O=141ml

b)m muối = x (g)

m FeSO4=\(15,2+\dfrac{152x}{278}=20\%\left(x+\dfrac{1520}{9}\right)\)

=>x=53,57g

Đáp án B

Ta có P₂O₅ + 3H₂O → 2H₃PO₄

n_H3PO4= 2.n_P2O5=0,03 mol, n_NaOH=0,04 mol

Ta xét tỉ lệ T= n_NaOH/ n_H3PO4= 0,04/0,03= 1,333

→ 1 < T < 2

→ Khi H₃PO₄ tác dụng với NaOH xảy ra 2 phương trình sau:

H₃PO₄ + NaOH → NaH₂PO₄ + H₂O

x              x                  x mol

H₃PO₄ + 2NaOH → Na₂HPO₄ + 2 H₂O

y               2y                y mol

ta có: n_H3PO4 = x+ y= 0,03 ; n_NaOH= x+2y= 0,04

Suy ra x= 0,02 ; y= 0,01

→C_{M NaH2PO4}= 0,02/0,4= 0,05M ;

C_{M Na2HPO4}= 0,01/0,4= 0,025M

a, \(C\%_{KCl}=\dfrac{20}{20+60}.100\%=25\%\)

b, \(C\%=\dfrac{40}{40+150}.100\%\approx21,05\%\)

c, \(C\%_{NaOH}=\dfrac{60}{60+240}.100\%=20\%\)

d, \(C\%_{NaNO_3}=\dfrac{30}{30+90}.100\%=25\%\)

e, \(m_{NaCl}=150.60\%=90\left(g\right)\)

f, \(m_{ddA}=\dfrac{25}{10\%}=250\left(g\right)\)

g, \(n_{NaOH}=120.20\%=24\left(g\right)\)

Gọi: n_{NaOH (thêm vào)} = a (g)

\(\Rightarrow\dfrac{a+24}{a+120}.100\%=25\%\Rightarrow a=8\left(g\right)\)

`a)m_[dd A]=[50,5.100]/10=505(g)`

`m_[H_2 O]=505-50,5=454,5(g)`

`b)n_[KNO_3]=[50,5]/101=0,5(mol)`

   `V_[dd B]=[0,5]/2=0,25(l)`

`c)20=[m+50,5]/[m+505].100`

`<=>m=63,125(g)`