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\(BaO+H_2SO_4->BaSO_4+H_2O\\ BaO+H_2O->Ba\left(OH\right)_2\\ n_{BaO}=\dfrac{30,6}{153}=0,2mol\\ n_{H_2SO_4}=0,098\cdot\dfrac{50}{98}=0,05mol\\ BaOdư\left(0,15mol=n_{Ba\left(OH\right)_2}\right)\\ C\%_{Ba\left(OH\right)_2}=\dfrac{0,15.171}{30,6+50-233.0,05}.100\%=37,2\%\)
\(BaO+H_2SO_4->BaSO_4+H_2O\\ n_{BaO}=\dfrac{30,6}{153}=0,2mol\\ n_{H_2SO_4}=0,098\cdot\dfrac{50}{98}=0,05mol\)
Vì acid hết, BaO dư nên C% dung dịch sau bằng 0%
\(n_{NaOH}=1.0,5=0,5(mol)\\ 2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ \Rightarrow n_{H_2SO_4}=0,25(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,25.98}{9,8\%}=250(g)\)
Ta có: \(C_{\%_{Ba\left(OH\right)_2}}=\dfrac{m_{Ba\left(OH\right)_2}}{250}.100\%=34,2\%\)
=> \(m_{Ba\left(OH\right)_2}=85,5\left(g\right)\)
=> \(n_{Ba\left(OH\right)_2}=\dfrac{85,5}{171}=0,5\left(mol\right)\)
Ta lại có: \(C_{\%_{H_2SO_4}}=\dfrac{m_{H_2SO_4}}{150}.100\%=4,9\%\)
=> \(m_{H_2SO_4}=7,35\left(g\right)\)
=> \(n_{H_2SO_4}=\dfrac{7,35}{98}=0,075\left(mol\right)\)
a. PTHH; Ba(OH)2 + H2SO4 ---> BaSO4↓ + 2H2O
Ta thấy: \(\dfrac{0,5}{1}>\dfrac{0,075}{1}\)
Vậy Ba(OH)2 dư.
Theo PT: \(n_{BaSO_4}=n_{H_2SO_4}=0,075\left(mol\right)\)
=> \(m_{BaSO_4}=0,075.233=17,475\left(g\right)\)
b. Ta có: \(m_{dd_{BaSO_4}}=250+7,35=257,35\left(g\right)\)
=> \(C_{\%_{BaSO_4}}=\dfrac{17,475}{257,35}.100\%=6,79\%\)
\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)
1 1 1 1
0,3 0,3 0,3
a) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)
b) \(n_{H2SO4}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(m_{H2SO4}=0,3.98=29,4\left(g\right)\)
\(m_{ddH2SO4}=\dfrac{29,4.100}{9,8}=300\left(g\right)\)
Chúc bạn học tốt
\(n_{FeO}=\dfrac{14,4}{72}=0,2\left(mol\right)\)
PTHH : \(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)
Theo Pt : \(n_{FeO}=n_{H2SO4}=n_{FeSO4}=0,2\left(mol\right)\)
\(m_{ddH2SO4}=\dfrac{0,2.98}{9,8\%}.100\%=200\left(g\right)\)
\(m_{ddspu_{ }}=14,4+200=214,4\left(g\right)\)
\(C\%_{FeSO4}=\dfrac{0,2.152}{214,4}.100\%=14,18\%\)
\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2SO_4}=n_{H_2}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.0,3=0,15\left(mol\right)\\ m=m_{ddH_2SO_4}=\dfrac{0,45.98.100}{9,8}=450\left(g\right)\\ a=m_{Al_2\left(SO_4\right)_3}=342.0,15=51,3\left(g\right)\\ V=V_{H_2\left(đktc\right)}=0,45.22,4=10,08\left(l\right)\)
a) \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
\(n_{KOH}=\dfrac{200.11,2\%}{56}=0,4\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{1}{2}n_{KOH}=0,2\left(mol\right)\)
\(m_{ddH_2SO_4}=\dfrac{0,2.98}{10\%}=196\left(g\right)\)
b) \(n_{K_2SO_4}=\dfrac{1}{2}n_{KOH}=0,2\left(mol\right)\)
\(m_{ddsaupu}=200+196=396\left(g\right)\)
=> \(C\%_{K2SO4}=\dfrac{0,2.174}{396}.100=8,79\%\)
c) \(3KOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3KCl\)
\(n_{FeCl_3}=n_{Fe\left(OH\right)_3}=\dfrac{1}{3}n_{KOH}=\dfrac{2}{15}\left(mol\right)\)
=>\(V_{FeCl_3}=\dfrac{2}{15}=0,13\left(l\right)\)
\(m_{Fe\left(OH\right)_3}=\dfrac{2}{15}.107=14,27\left(g\right)\)
\(BaO+H_2SO_4->BaSO_4+H_2O\\ n_{BaO}=\dfrac{15,3}{153}=0,1mol\\ n_{H_2SO_4}=0,098\cdot\dfrac{250}{98}=0,25mol\\ H_2SO_4:dư\left(0,15mol\right)\\ C_{\%H_2SO_4dư}=\dfrac{0,15.98}{15,3+250-233.0,1}.100\%=6,07\%\)
C% dung dịch thu được mà bạn