a) C_{M HCl} = \(\dfrac{10\times7,3\times1,2}{36,5}=2,4M\)

b) nHCl = 2,4 . 0,2 = 0,48 mol

Pt: NaOH + HCl --> NaCl + H₂O

..0,48 mol<-0,48 mol

V_NaOH = \(\dfrac{0,48}{0,2}=2,4M\)

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)

Bài 1

\(a,n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ CuO+2HCl\xrightarrow[]{}CuCl_2+H_2O\\ n_{CuCl_2}=n_{CuO}=0,2mol\\ m_{CuCl_2}=0,2.135=27\left(g\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ C_{MHCl}=\dfrac{0,4}{0,5}=0,8\left(M\right)\)

Bài 5

\(a,n_{NaOH}=0,2.1=0,2\left(mol\right)\\ 2NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ n_{H_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MH_2SO_4}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\ b,n_{Na_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MNa_2SO_4}=\dfrac{0,1}{0,2+0,4}=\dfrac{1}{6}\left(M\right)\\ c,m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)

200ml = 0,2l

\(n_{Ba\left(OH\right)2}=0,5.0,2=0,1\left(mol\right)\)

Pt : \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O|\)

              1              2             1            2

             0,1           0,2          0,1

a) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

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

b) \(n_{BaCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{BaCl2}=0,1.208=20,8\left(g\right)\)

c) \(V_{ddspu}=0,2+0,2=0,4\left(l\right)\)

\(C_{M_{BaCl2}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)

 Chúc bạn học tốt

PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Ta có: \(n_{Ba\left(OH\right)_2}=0,2\cdot0,5=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{BaCl_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\\m_{BaCl_2}=0,1\cdot208=20,8\left(g\right)\\C_{M_{BaCl_2}}=\dfrac{0,1}{0,2+0,2}=0,25\left(M\right)\end{matrix}\right.\)

Có \(m_M=m_{Na}+m_{Cl}+m_{SO_4}=1,38+m_{Cl}+m_{SO_4}=3,76\)

\(\Rightarrow m_{Cl}+m_{SO_4}=2,38\)

\(\Rightarrow35,5n_{HCl}+96n_{H_2SO_4}=2,38\)

Lại có : \(n_O=n_{NaOH}=0,06\left(mol\right)\)

BtH : \(n_H=2n_{H_2O}=2n_O=n_{H\left(NaOH\right)}+n_{H\left(HCl\right)}+2n_{H\left(H_2SO_4\right)}\)

\(\Rightarrow n_{HCl}+2n_{H_2SO_4}=0,06\)

\(\Rightarrow n_{HCl}=0,04\left(mol\right)\)

\(\Rightarrow C_{MHCl}=1M\)

PTHH : HCl+NaOH-----> NaCl +H2O

................0,3...0,3(mol)

\(C_{M\left(NaOH\right)}=\frac{0,3}{0,1}=3\left(M\right)\)

#Walker

\(n_{HCl}=2.0,15=0,3\left(mol\right)\\ PTHH:NaOH+HCl\rightarrow NaCl+H_2O\\ C_{M_{NaOH}}=\frac{0,3}{0,1}=3\left(M\right)\)

#Alec

PTHH : NaOH + HCl--->NaCl +H2O

n HCl= 0,15.2 = 0,3 mol

n NaOH= nHCl = 0,3 mol

=> C_{M NaOH} =0,3/ 0,1 = 3M