CHo m(g) dd H2SO4 10% trung hòa vừa đủ 160g dd NaOH 20% sau phản ứng thu đc dung dịch X
a, tìm m
b, tìm nồng độ % của dd X
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B1:
2NaOH+H2SO4\(\rightarrow\)Na2SO4+2H2O
nNaOH=\(\frac{4}{40}=0.1\)mol
=>nH2SO4=\(\frac{1}{2}\)nNaOH=0.05 mol
=>CM=\(\frac{n_{H2SO42}}{V}\)=\(\frac{0.05}{200}\)=2,5.10-4 (M)
B2:
Mg+\(\frac{1}{2}\)O2\(\underrightarrow{t^0}\)MgO (1)
MgO+2HCl\(\rightarrow\)MgCl2+H2O (2)
nMg(1)=\(\frac{0,36}{24}=0,015mol\)
=>nMgO(1)=0,015=nMgO(2)
nHCl(2)=2nMgO(2)=0,03mol
=>CM(HCl)=\(\frac{n_{HCl}}{V}=\frac{0,03}{100}=3.10^{-4}M\)
\(n_{NaOH}=0,8\left(mol\right)\)
\(PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
\(n_{H2SO4}=\frac{1}{2}n_{NaOH}=0,4\left(mol\right)\)
\(\rightarrow m_{H2SO4}=0,4.98=39,2\left(g\right)\)
\(m_{dd_{H2SO4}}=392\left(g\right)\)
\(m_{dd_{spu}}=160+392=552\left(g\right)\)
\(\rightarrow C\%_{dd_X}=10,29\%\)
\(PTHH:H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
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
=> \(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\)
=> \(m_{ddH_2SO_4}=\dfrac{44,1.100}{20}=220,5\left(g\right)\)
mdd (20oC) = 10,8 + 220,5 - 0,45.2 - 14,76 = 215,64 (g)
\(m_{MgSO_4\left(dd.ở.20^oC\right)}=\dfrac{215,64.21,703}{100}=46,8\left(g\right)\)
=> nMgSO4 (tách ra) = \(0,45-\dfrac{46,8}{120}=0,06\left(mol\right)\)
=> nH2O (tách ra) = \(\dfrac{14,76-0,06.120}{18}=0,42\left(mol\right)\)
Xét nMgSO4 (tách ra) : nH2O (tách ra) = 0,06 : 0,42 = 1 : 7
=> CTHH: MgSO4.7H2O
\(n_{NaOH}=\frac{160.20}{100.40}=0,8\left(mol\right)\\ PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ m_{H_2SO_4}=\left(\frac{0,8}{2}\right).98=39,2\left(g\right)\\ m_{ddH_2SO_4}=\frac{39,2.100}{10}=392\left(g\right)\\ m_{ddspu}=160+392=552\left(g\right)\\ C\%_{ddX}=\frac{0,4.142}{552}.100\%=10,29\left(\%\right)\)