\(n_{SO_3}=\dfrac{32}{80}=0.4\left(mol\right)\)

\(m_{H_2SO_4}=200\cdot10\%=20\left(g\right)\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(0.4.....................0.4\)

\(m_{dd}=32+200=232\left(g\right)\)

\(C\%H_2SO_4=\dfrac{0.4\cdot98+20}{232}\cdot100\%=25.57\%\)

có ai giúp mình ko ạ

m H2SO4 = 200.10% = 20(gam)

n SO3 = 32/80 = 0,4(mol)

SO3 + H2O → H2SO4

0,4.......0,4..........0,4........(mol)

Sau phản ứng :

m H2SO4 = 20 + 0,4.98 = 59,2(gam)

m dd = 32 + 200 - 0,4.18 = 224,8(gam)

C% H2SO4 = 59,2/224,8  .100% = 26,33%

\(1.SO_3+H_2O\rightarrow H_2SO_4\\ 2.m_{H_2SO_4}=\dfrac{200.19,6\%}{100\%}=39,2g\\ n_{H_2SO_4}=\dfrac{39,2}{98}=0,4mol\\ n_{SO_2}=n_{H_2SO_4}=0,4mol\\ m=m_{SO_2}=0,4.64=25,6g\)

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

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)

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

\(SO_3+H_2O\rightarrow H_2SO_4\)

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

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

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

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1mol\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

0,1             0,3                 0,1              0,3

\(m_{H_2SO_4}=0,3\cdot98=29,4\left(g\right)\)\(\Rightarrow m_{ddH_2SO_4}=\dfrac{29,4}{25}\cdot100=117,6\left(g\right)\)

\(m_{H_2O}=0,3\cdot18=5,4\left(g\right)\)

\(m_{Fe_2\left(SO_4\right)_3}=0,1\cdot400=40\left(g\right)\)

\(m_{ddsau}=16+117,6-5,4=128,2\left(g\right)\)

\(C\%=\dfrac{40}{128,2}\cdot100\%=31,2\%\)

Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ ---> Fe₂(SO₄)₃ + 3H₂O

Theo PT: \(n_{H_2SO_4}=3.n_{H_2SO_4}=3.0,1=0,3\left(mol\right)\)

=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{29,4}{m_{dd_{H_2SO_4}}}.100\%=25\%\)

=> \(m_{dd_{H_2SO_4}}=117,6\left(g\right)\)

=> \(m_{dd_{Fe_2\left(SO_4\right)_3}}=117,6+16=133,6\left(g\right)\)

Theo PT: \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,1\left(mol\right)\)

=> \(m_{Fe_2\left(SO_4\right)_3}=0,1.400=40\left(g\right)\)

=> \(C_{\%_{Fe_2\left(SO_4\right)_3}}\dfrac{40}{133,6}.100\%=29,94\%\)

a) PTHH: \(SO_3+H_2O\rightarrow H_2SO_4\)

Ta có: \(n_{SO_3}=\dfrac{24}{80}=0,3\left(mol\right)=n_{H_2SO_4}\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,3\cdot98}{20\%}=147\left(g\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{147}{1,14}\approx128,95\left(ml\right)\)

b) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

Theo PTHH: \(n_{Fe}=n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)=n_{FeSO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,3\cdot56=16,8\left(g\right)\\V_{H_2}=0,3\cdot24,76=7,428\left(l\right)\\m_{FeSO_4}=0,3\cdot152=45,6\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}=163,2\left(g\right)\)

\(\Rightarrow C\%_{FeSO_4}=\dfrac{45,6}{163,2}\cdot100\%\approx27,94\%\)

\(a,C_{M\left(NaOH\right)}=\dfrac{0,3}{0,5}=0,6M\\ b,n_{NaOH}=\dfrac{24}{40}=0,6\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,6}{0,4}=1,5M\)

`a)C%_[KOH]=28/140 . 100=20%`

`b)C%_[KOH]=80/[80+320] .100=20%`

\(a,C\%_{KOH}=\dfrac{28}{140}.100\%=20\%\\ b,C\%_{KOH}=\dfrac{80}{80+320}.100\%=20\%\)

Đổi 500 ml = 0,5 (l)

Số mol SO3 là: 24 : 80 = 0,3 (mol) 

SO3 + H2O = H2SO4

0,3                       0,3     (mol)

Nồng độ mol dung dịch là : 0,3 : 0,5 = 0,6 (M)

Al + H2SO4 = Al2(SO4)3 + H2

0,3      0,3                           (mol)

Khối lượng Al cần tìm là: 27 x 0,3 = 8,1 (g)