\(a,n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\\ C_{M\left(HCl\right)}=\dfrac{0,2}{0,4}=0,5M\\ b,n_{H_2SO_4}=\dfrac{73,5}{98}=0,75\left(mol\right)\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,75}{0,5}=1,5M\\ n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,4}{0,25}=1,6M\\ n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\\ C_{M\left(Ba\left(OH\right)_2\right)}=\dfrac{0,2}{0,8}=0,25M\)

\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\\ PTHH:Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=2.0,25=0,5\left(mol\right)\\ a,C_{MddNaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\\ b,2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ n_{H_2SO_4}=n_{Na_2SO_4}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{H_2SO_4}=0,25.98=24,5\left(g\right)\\ m_{ddH_2SO_4}=\dfrac{24,5.100}{20}=122,5\left(g\right)\\ V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,456\left(ml\right)\\ c,V_{ddsau}=V_{ddNaOH}+V_{ddH_2SO_4}\approx0,5+0,107456=0,607456\left(l\right)\\C_{MddNa_2SO_4}\approx\dfrac{ 0,25}{0,607456}\approx0,411552\left(M\right)\)

a)

$BaO + H_2O \to Ba(OH)_2$
$n_{Ba(OH)_2} = n_{BaO} = \dfrac{30,6}{153} = 0,2(mol)$
$C_{M_{Ba(OH)_2}} = \dfrac{0,2}{0,2} = 1M$

b)

$Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
$n_{HCl} = 2n_{Ba(OH)_2} = 0,4(mol)$
$V_{dd\ HCl} = \dfrac{0,4}{0,5} = 0,8(lít)$

Đặt x, y là số mol HCl và  H 2 SO 4  trong 40 ml dung dịch A.

HCl + NaOH → NaCl +  H 2 O

H 2 SO 4 + 2NaOH →  Na 2 SO 4  + 2 H 2 O

Số mol NaOH: x + 2y = 1.60/1000 = 0,06 mol (1)

Khối lượng 2 muối : 58,5x + 142y = 3,76 (2)

Từ (1), (2), giải ra : x = 0,04 ; y = 0,01.

C M HCl  = 0,04/0,04 = 1(mol/l)

C M H 2 SO 4 = 0,01/0,04 = 0,25 (mol/l)

Bài1:

a,Vì dd A là dd bazo nên làm cho quỳ tím đổi thành màu xanh

b,\(n_{Na_2O}=\dfrac{21,7}{62}=0,35\left(mol\right)\)

PTHH: Na₂O + H₂O → 2NaOH

Mol:       0,35                     0,7

\(\Rightarrow C_{M_{ddNaOH}}=\dfrac{0,7}{0,4}=1,75M\)

Bài 2:

a,\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:    0,15     0,3                      0,15

⇒ a=m_Zn = 0,15.65 = 9,75 (g)

b,\(V_{HCl}=\dfrac{0,3}{1,5}=0,2\left(l\right)=200\left(ml\right)\)

a) $n_{NaOH} = \dfrac{15,5}{40} = 0,3875(mol)$
$C_{M_{NaOH}} = \dfrac{0,3875}{0,5} =0,775M$
b)

$2NaOH + H_2SO_4 \to Na_2SO_4 + H_2$

$n_{H_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,19375(mol)$
$m_{dd\ H_2SO_4} =\dfrac{0,19375.98}{20\%} = 94,9375(gam)$
$V_{dd\ H_2SO_4} = \dfrac{94,9375}{1,14} = 83,28(ml)$

\(C_{M\left(H_2SO_4\right)}=a;C_{M\left(NaOH\right)}=b\\ H_2SO_4+2NaOH->Na_2SO_4+2H_2O\\ 0,015a\cdot2-0,036b=0\\ Ba\left(OH\right)_2+H_2SO_4->BaSO_4+2H_2O\\ 0,04a=\dfrac{0,056b}{2}+\dfrac{0,466}{233}=0,028b+0,002\\ a=0,12M;b=0,1M\)

\(1,C_{M\left(HCl\right)}=\dfrac{0,75}{0,5}=1,5M\\ 2,n_{Ca\left(OH\right)_2}=\dfrac{37}{74}=0,5\left(mol\right)\\ C_{M\left(Ca\left(OH\right)_2\right)}=\dfrac{0,5}{1,5}=0,33M\\ 3,n_{NaOH}=0,25+\dfrac{20}{40}=0,75\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,75}{2}=0,375M\\ 4,n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,5}{2}=0,25M\)

`1) C_[M_[HCl]] = [ 0,75 ] / [ 0,5 ] = 1,5 (M)`

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`2)n_[Ca(OH)_2] = 37 / 74 = 0,5 (mol)`

`-> C_[M_[Ca(OH)_2]] = [ 0,5 ] / [ 1,5 ] ~~ 0,33 (M)`

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`3) n_[NaOH] = 0,25 + 20 / 40 = 0,75 (mol)`

`-> C_[M_[NaOH]] = [ 0,75 ] / 2 = 0,375 (M)`

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`4) n_[H_2 SO_4] = 49 / 98 = 0,5 (mol)`

`-> C_[M_[H_2 SO_4]] = [ 0,5 ] / 2 = 0,25 (M)`