\(n_{SO_3}=\dfrac{8}{80}=0,1\left(mol\right)\)

PTHH: SO₃ + H₂O --> H₂SO₄

             0,1------------->0,1

\(m_{H_2SO_4\left(bđ\right)}=242.10\%=24,2\left(g\right)\)

m_{H2SO4(sau pư)} = 24,2 + 0,1.98 = 34 (g)

m_{dd sau pư} = 8 + 242 = 250 (g)

\(C\%_{dd.H_2SO_4.sau.pư}=\dfrac{34}{250}.100\%=13,6\%\)

\(a.Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ b.n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\\ n_{H_2SO_4}=3.0,15=0,45\left(mol\right)\\ m_{ddH_2SO_4}=\dfrac{0,45.98.100}{14,7}=300\left(g\right)\\ m_{ddsau}=24+300=324\left(g\right)\\ n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,15\left(mol\right)\\ C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{0,15.400}{324}.100\approx18,519\%\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(n_{SO_3}=\dfrac{12}{80}=0,15\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{SO_3}=0,15\left(mol\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,15.98}{12+100}.100\%=13,125\%\)

\(SO_3+H_2O\rightarrow H_2SO_4\\ m_{H_2SO_4\left(tăng\right)}=\dfrac{98}{80}m=\dfrac{49}{40}m=1,225m\left(g\right)\\ m_{H_2SO_4\left(dd.14,7\%\right)}=14,7.200=29,4\left(g\right)\\ Ta.có:C\%_{ddH_2SO_4\left(cuối\right)}=20\%\\ \Leftrightarrow\dfrac{29,4+1,225m}{m+200}.100\%=20\%\\ m\approx10,3415\left(g\right)\)

Câu 1 : 

Gọi $n_{SO_3} = a(mol)$

$SO_3 + H_2O \to H_2SO_4$

Sau khi pha : 

$m_{dd} = 80a + 200(gam)$
$m_{H_2SO_4} = 98a + 200.9,8\% = 98a + 19,6(gam)$

Suy ra : $C\%_{H_2SO_4} = \dfrac{98a + 19,6}{200 + 80a}.100\% = 49\%$
$\Rightarrow a = 1,333(mol)$
$m_{SO_3} = 1,333.80 = 106,64(gam)$

Gọi $m_{oleum} = a(gam) ;  m_{H_2SO_4} = b(gam)$
Ta có : 

Sau khi trộn  :

$m_{oleum} = a + b(gam)$
$m_{SO_3} = a.71\% = 0,71a(gam)$

$\Rightarrow \%SO_3 = \dfrac{0,71a}{a + b}.100\% = 62\%$

$\Rightarrow a + b = 0,4402a$

$\Rightarrow \dfrac{a}{b} = \dfrac{1}{1 - 0,4402} = 1,78$

1.

\(m_{H_2SO_4\left(9.8\%\right)}=200\cdot9.8\%=19.6\left(g\right)\)

\(n_{H_2SO_4}=\dfrac{19.6}{98}=0.2\left(mol\right)\)

\(TC:n_{SO_3}=a\left(mol\right)\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(a................a\)

\(m_{H_2SO_4\left(tt\right)}=98a\left(g\right)\)

\(m_{H_2SO_4\left(tổng\right)}=19.6+98a\left(g\right)\)

\(m_{dd_{H_2SO_4}}=200+80a\left(g\right)\)

\(C\%_{H_2SO_4}=\dfrac{19.6+98a}{200+80a}\cdot100\%=49\%\)

\(\Rightarrow a=\dfrac{4}{3}\)

\(m_{SO_3}=\dfrac{4}{3}\cdot80=106.67\left(g\right)\)

\(m_{H_2SO_4}=9.8\left(g\right)\)

\(n_{H_2SO_4}=\dfrac{9.8}{98}=0.1\left(mol\right)\)

\(Đặt:n_{Ba\left(1\right)}=a\left(mol\right)\)

\(Ba+H_2SO_4\rightarrow BaSO_4+H_2\)

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

\(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\left(1\right)\)

\(a..................a........a\)

\(m_{Ba\left(OH\right)_2}=171a\left(g\right)\)

\(m_{dd}=m_{Ba}+m_{ddH_2SO_4}-m_{H_2}-m_{BaSO_4}=\left(0.1+a\right)\cdot137+200-23.3-\left(0.1+a\right)\cdot2=190.2+136.8a\left(g\right)\)

\(C\%Ba\left(OH\right)_2=\dfrac{171a}{190.2+136.8a}\cdot100\%=2.51\%\)

\(\Leftrightarrow a=0.028\)

\(m_{Ba}=\left(0.1+0.028\right)\cdot137=17.536\left(g\right)\)

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