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

Bạn ơi , câu 1 sao dd NaCl lại tác dụng với NaCl v ?

\(n_{NaOH}=0,8\left(mol\right)\)

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

\(n_{H2SO4}=\frac{1}{2}n_{NaOH}=0,4\left(mol\right)\)

\(\rightarrow m_{H2SO4}=0,4.98=39,2\left(g\right)\)

\(m_{dd_{H2SO4}}=392\left(g\right)\)

\(m_{dd_{spu}}=160+392=552\left(g\right)\)

\(\rightarrow C\%_{dd_X}=10,29\%\)

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

Ban đầu 0,1________0,8

Phản ứng 0,1_________ 0,2 ______0,1

Dư ______ 0 ______ 0,6

\(m_{NaOH}=160.20\%=32g\)

\(n_{NaOH}=\frac{32}{23+17}=0,8\left(mol\right)\)

\(m_{H2SO4}=200.4,9\%=9,8\%\)

\(n_{H2SO4}=\frac{9,8}{32+2+16.4}=0,1\left(mol\right)\)

\(C\%_{Na2SO4}=\frac{0,1.\left(23.2+32+16.4\right)}{200+160}.100\%=3,94\%\)

\(C\%_{NaOH_{Du}}=\frac{0,6.\left(23+17\right)}{200}.100\%=65,16\%\)

\(n_{FeCl_3}=\dfrac{48,75}{162,5}=0,3(mol)\\ 3NaOH+FeCl_3\to Fe(OH)_3\downarrow+3NaCl\\ \Rightarrow n_{Fe(OH)_3}=0,3(mol);n_{NaOH}=n_{NaCl}=0,9(mol)\\ a,m_{Fe(OH)_3}=0,3.107=32,1(g)\\ b,m_{dd_{NaOH}}=\dfrac{0,9.40}{10\%}=360(g)\\ c,C\%_{NaCl}=\dfrac{0,9.58,5}{360+48,75-32,1}.100\%=13,98\%\\ \)

\(d,2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ \Rightarrow n_{H_2SO_4}=0,45(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,45.98}{20\%}=220,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{220,5}{1,14}=193,42(ml)\)

\(n_{Mg}=\dfrac{10,8}{24}=0,45\left(mol\right)\)

PTHH: Mg + H₂SO₄ --> MgSO₄ + H₂

         0,45-->0,45------>0,45--->0,45

=> \(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\)

=> \(m_{ddH_2SO_4}=\dfrac{44,1.100}{20}=220,5\left(g\right)\)

m_{dd (20^oC)} = 10,8 + 220,5 - 0,45.2 - 14,76 = 215,64 (g)

\(m_{MgSO_4\left(dd.ở.20^oC\right)}=\dfrac{215,64.21,703}{100}=46,8\left(g\right)\)

=> n_{MgSO4 (tách ra)} = \(0,45-\dfrac{46,8}{120}=0,06\left(mol\right)\)

=> n_{H2O (tách ra)} = \(\dfrac{14,76-0,06.120}{18}=0,42\left(mol\right)\)

Xét n_{MgSO4 (tách ra)} : n_{H2O (tách ra)} = 0,06 : 0,42 = 1 : 7

=> CTHH: MgSO₄.7H₂O