Chọn B

Nồng độ mol của dung dịch  H 2 SO 4  :

n H 2 SO 4 = n SO 3  = 8/80 = 0,1 mol

Theo phương trình hoá học :

C Mdd   H 2 SO 4  = 1000x0,1/250 = 0,4M

1. \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

PTHH: Na₂O + H₂O → 2NaOH

Mol:      0,25                     0,5

\(C_{M_{ddNaOH}}=\dfrac{0,5}{0,5}=1M\)

2. 

PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mol:        0,5           0,25

\(m_{ddH_2SO_4}=\dfrac{0,25.98.100}{20}=122,5\left(g\right)\)

\(V_{ddH_2SO_4}=\dfrac{122,5}{1,14}=107,456\left(ml\right)\)

\(n_{Na2O}=\dfrac{m_{Na2O}}{M_{Na2O}}=0,25\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,25 mol - 0,25 mol - 0,5 mol

a) \(C_{M_{NaOH}}=\dfrac{n_{NaOH}}{V_{NaOH}}=1\left(M\right)\)

b) \(H_2SO_4+2NaOH\rightarrow Na_2SO4+2H_2O\)

0,25 mol - 0,5 mol - 0,25 mol - 0,5 mol

\(m_{ctH2SO4}=n_{H2SO4}.M_{H2SO4}=24,5\left(g\right)\)

\(C_{\%_{H2SO4}}=\dfrac{m_{ctH2SO4}}{m_{ddH2SO4}}.100\%\)

\(\Rightarrow m_{ddH2SO4}=\dfrac{m_{ctH2SO4}.100\%}{C_{\%_{H2SO4}}}=122,5\left(g\right)\)

\(D_{H2SO4}=\dfrac{m_{ddH2SO4}}{V_{H2SO4}}\Rightarrow V_{H2SO4}=\dfrac{m_{ddH2SO4}}{D_{H2SO4}}\approx107,46\left(ml\right)\)

mHCl =( c% .mdd) /100 = (3,65 .400) /100 =14,6gam 

nHCl = 14,6/36,5 =0.4 mol

CM =  n/v = 0,4/ 21 =~0,02

0,1M

0, 1M

n_CuO=16/80=0,2mol

CuO+H₂SO₄ --->CuSO₄+H₂O

0,2mol-->0,2mol -->0,2mol

C_MH2SO4=0,2/0,5=0,4M

m_CuSO4=0,2.160=32g

C_MCuSO4=0,2/0,5=0,4M

n_CuO = \(\dfrac{16}{80}\)= 0,2 mol

V = 500 ml = 0,5 (l)

CuO + H₂SO₄ -> CuSO_{4 (x)} + H₂O

0,2--->0,2 mol--->0,2mol

a) C_M(H2SO4) = \(\dfrac{0,2}{0,5}\)=0,4 M

b) m_CuSO4 = 0,2 . 160 = 32 g

\(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)

PTHH:

\(Na_2O+H_2O\rightarrow2NaOH\)

0,2          0,2          0,4

\(C_{M\left(NaOH\right)}=\dfrac{0,4}{0,5}=0,8\left(M\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

0,4           0,2

\(V_{H_2SO_4}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)

\(a)Na_2O+H_2O\rightarrow2NaOH\\ n_{Na_2O}=\dfrac{12,4}{62}=0,2mol\\ n_{NaOH}=0,2.2=0,4mol\\ C_{M_{NaOH}}=\dfrac{0,4}{0,5}=0,8M\\ b)2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ n_{H_2SO_4}=0,4:2=0,2mol\\ V_{ddH_2SO_4}=\dfrac{0,2}{1}=0,2l=200ml\)

\(n_{H2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

 Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)

        1           1                1           1

      0,25     0,25                           0,25

a) \(n_{Fe}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)

⇒ \(m_{Fe}=0,25.56=14\left(g\right)\)

b) \(n_{H2SO4}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)

200ml = 0,2l

\(C_{M_{ddH2SO4}}=\dfrac{0,25}{0,2}=1,25\left(M\right)\)

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