a)

$m_{dd} = 16 + 234 = 250(gam)$
$V_{dd} = \dfrac{250}{1,05} = 238(ml)$

$n_{NaOH} = \dfrac{16}{40} = 0,4(mol)$

Suy ra :

$C\%_{NaOH} = \dfrac{16}{250} = 6,4\%$

$C_{M_{NaOH}} = \dfrac{0,4}{0,238} = 1,68M$

b)

$C\%_{NaOH} = \dfrac{16+10}{250+10}.100\% = 10\%$

\(n_{Na_2O}=\dfrac{6.2}{62}=0.1\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(0.1........................0.2\)

\(m_{NaOH}=0.2\cdot40=8\left(g\right)\)

\(m_{dd}=6.2+2000=2006.2\left(g\right)\)

\(C\%_{NaOH}=\dfrac{8}{2006.2}\cdot100\%=0.4\%\)

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

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{Fe}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(n_{NaOH}=2n_{Na_2O}=0,2\left(mol\right)\Rightarrow C_{M_{NaOH}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)

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

PTHH: 

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

0,15    0,3           0,15      0,15

\(m_{Fe}=0,15.56=8,4\left(g\right)\)

\(a,n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

PTHH :

\(Na_2O+H_2O\rightarrow2NaOH\)

0,1           0,1        0,2

\(C_{M\left(A\right)}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)

a) \(m_{NaOH}=\dfrac{60.20}{100}=12\left(g\right)\)

\(C\%_{dd.sau.khi.pha}=\dfrac{12}{60+40}.100\%=12\%\)

b) \(C\%_{dd.sau.khi.pha}=\dfrac{12+12}{60+12}.100\%=33,33\%\)

60g dd 20% có 60.20%=12g NaOH, 60-12=48g H2O

a, Pha thêm 40g H2O , ta có 88g H2O

→C%NaOH=12.100:88=12,64% 

b) 

tan thêm 12 g

=>m NaOH=24g

=>C%=\(\dfrac{24}{60+12}100\)=33,33%

\(m_{NaOH}=\dfrac{60.20}{100}=12\left(g\right)\\ a,m_{\text{dd}}=60+40=100\left(g\right)\\ C\%=\dfrac{12}{100}.100\%=12\%\\ b,m_{NaOH}=12+12=24\left(g\right)\\ C\%=\dfrac{24}{60}.100\%=40\%\)

B1:

2NaOH+H₂SO₄\(\rightarrow\)Na₂SO₄+2H₂O

n_NaOH=\(\frac{4}{40}=0.1\)mol

=>n_H2SO4=\(\frac{1}{2}\)n_NaOH=0.05 mol

=>C_M=\(\frac{n_{H2SO42}}{V}\)=\(\frac{0.05}{200}\)=2,5.10^-4 (M)

B2:

Mg+\(\frac{1}{2}\)O₂\(\underrightarrow{t^0}\)MgO                (1)

MgO+2HCl\(\rightarrow\)MgCl₂+H₂O (2)

n_Mg(1)=\(\frac{0,36}{24}=0,015mol\)

=>n_MgO(1)=0,015=n_MgO(2)

n_HCl(2)=2n_MgO(2)=0,03mol

=>C_M(HCl)=\(\frac{n_{HCl}}{V}=\frac{0,03}{100}=3.10^{-4}M\)

nNa2O=0.1(mol)

pthh:Na2O+H2O->2NaOH

theo pthh:nH2O=nNa2O->nH2O=0.1(mol)->mH2O=0.1*18=1.8(g)

b)Theo pthh:nNaOH=2 nNa2O

->nNaOH=0.1*2=0.2(mol)

CM=0.2:0.2=1(M)

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

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