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\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)
PTHH: Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
______0,05------>0,15--------->0,05
=> mH2SO4 = 0,15.98 = 14,7(g)
=> \(C\%\left(H_2SO_4\right)=\dfrac{14,7}{100}.100\%=14,7\%\)
\(C\%\left(Fe_2\left(SO_4\right)_3\right)=\dfrac{0,05.400}{8+100}.100\%=18,52\%\)
PTHH: Fe2(SO4)3 + 6NaOH --> 2Fe(OH)3\(\downarrow\) + 3Na2SO4
________0,05----------------------->0,1
=> mFe(OH)3 = 0,1.107=10,7(g)
a/
\(n_{Na_2O}=\dfrac{9,3}{62}=0,15\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
0,15 0,3 (mol)
\(m_{NaOH}=0,3.40=12\left(g\right)\)
\(m_A=90,7+9,3=100\left(g\right)\)
\(C\%_{NaOH}=\dfrac{12}{100}.100\%=12\%\)
b/
m\(_{FeSO_4}=\dfrac{16.200}{100}=32\left(g\right)\)
\(\rightarrow m_{FeSO_4}=\dfrac{32}{152}=\dfrac{4}{19}\left(mol\right)\)
\(2NaOH+FeSO_4\rightarrow Na_2SO_4+Fe\left(OH\right)_2\downarrow\)
bđ: 0,3 \(\dfrac{4}{19}\) 0 0 (mol)
pư: 0,3 0,15 0,15 0,15 (mol)
dư: 0 \(\dfrac{23}{380}\) (mol)
\(m_{Fe\left(OH\right)_2}=0,15.90=13,5\left(g\right)\)
\(m_C=100+200-13,5=286,5\left(g\right)\)
\(m_{Na_2SO_4}=0,15.142=21,3\left(g\right)\)
\(\rightarrow C\%_{Na_2SO_4}=\dfrac{21,3}{286,5}.100\%\approx7,4\%\)
\(m_{FeSO_4\left(dư\right)}=\dfrac{23}{380}.152=9,2\left(g\right)\)
\(\rightarrow C\%_{FeSO_4\left(dư\right)}=\dfrac{9,2}{286,5}.100\%\approx3,2\%\)
\(n_{BaCl_2}=\dfrac{208.10\%}{208}=0,1\left(mol\right)\\ a,BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ 0,1............0,1..............0,1.............0,2\left(mol\right)\\ b,m_{ddH_2SO_4}=\dfrac{0,1.98.100}{8}=122,5\left(g\right)\\ c,m_{kt}=m_{BaSO_4}=0,1.233=23,3\left(g\right)\\ d,m_{ddsau}=208+122,5-23,3=307,2\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,2.36,5}{307,2}.100\approx2,376\%\)
PTHH: \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
a) Ta có: \(n_{CuSO_4}=\dfrac{320\cdot20\%}{160}=0,4\left(mol\right)=n_{Cu\left(OH\right)_2}\) \(\Rightarrow m_{Cu\left(OH\right)_2}=0,4\cdot98=39,2\left(g\right)\)
b) Theo PTHH: \(n_{NaOH}=2n_{CuSO_4}=0,8mol\) \(\Rightarrow m_{ddNaOH}=\dfrac{0,8\cdot40}{10\%}=320\left(g\right)\)
c) Theo PTHH: \(n_{Na_2SO_4}=n_{CuSO_4}=0,4mol\) \(\Rightarrow m_{Na_2SO_4}=0,4\cdot142=56,8\left(g\right)\)
Mặt khác: \(m_{dd}=m_{ddNaOH}+m_{ddCuSO_4}-m_{Cu\left(OH\right)_2}=600,8\left(g\right)\)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{56,8}{600,8}\cdot100\%\approx9,45\%\)
Ta thấy khi cho Br 2 vào dung dịch 2 muối S 4 + thì toàn bộ S 4 + sẽ bị oxi hoá lên S 6 + ( SO 4 2 - )do đó :
n SO 2 = n SO 4 2 - = 0,15
=> m BaSO 4 = 0,15.233 = 34,95g
nNa = 6.9 : 23 = 0.3 mol
4Na + O2 ->2 Na2O
mol : 0.3 -> 0.15
Na2O + H2O -> 2NaOH
mol : 0.15 -> 0.3
mdd = 0.15 x 62 + 140.7 = 150g
C% NaOH = 0.3x40: 150 x 100% = 8%
\(n_{Mg\left(NO_3\right)_2}=0,05.0,8=0,04mol\)
\(n_{NaOH}=\dfrac{50.8}{40.100}=0,1mol\)
Mg(NO3)2+2NaOH\(\rightarrow\)Mg(OH)2\(\downarrow\)+2NaNO3
-Tỉ lệ: \(\dfrac{0,04}{1}< \dfrac{0,1}{2}\rightarrow\)NaOH dư, tính theo Mg(NO3)2
Mg(NO3)2+2NaOH\(\rightarrow\)Mg(OH)2\(\downarrow\)+2NaNO3
0,04\(\rightarrow\)..........0,08...........0,04.............0,08
\(n_{NaOH\left(dư\right)}=0,1-0,08=0,02mol\rightarrow m_{NaOH}=0,02.40=0,8gam\)
\(m_{Mg\left(OH\right)_2}=0,04.58=2,32gam\)
\(m_{dd_{Mg\left(NO_3\right)_2}}=v.d=50.1,14=57gam\)
\(m_{dd}=57+50-2,32=104,68gam\)
C%NaNO3=\(\dfrac{0,08.85}{104,68}.100\approx6,5\%\)
C%NaOH=\(\dfrac{0,8.100}{104,68}\approx0,76\%\)
Mg(OH)2\(\overset{t^0}{\rightarrow}\)MgO+H2O
\(n_{MgO}=n_{Mg\left(OH\right)_2\left(pu\right)}=0,04.\dfrac{75}{100}=0,03mol\)
\(n_{Mg\left(OH\right)_2\left(dư\right)}=0,04-0,03=0,01mol\)
-Chất rắn thu được gồm: Mg(OH)2 dư 0,01 mol và MgO 0,03mol
mrắn=0,01.58+0,03.40=1,78gam