\(SO_3+H_2O\rightarrow H_2SO_4\)

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

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_K = 

Pt: 2K + 2H₂O --> 2KOH + H₂

1 mol--------------------------> 0,5 mol

m_H2 = 0,5 . 2 = 1 (g)

mdd = m_K + m_nước - m_H2 = 39 + 362 - 1 = 400 (g)

C% dd KOH = 

https://hoc24.vn/hoi-dap/tim-kiem?id=562460&q=t%C3%ADnh%20n%E1%BB%93ng%20%C4%91%E1%BB%99%20ph%E1%BA%A7n%20tr%C4%83m%20c%E1%BB%A7a%20dung%20d%E1%BB%8Bch%20t%E1%BB%8Da%20th%C3%A0nh%20khi%20h%C3%B2a%20tan%20%3A%20%201%2F%2039g%20Kali%20v%C3%A0o%20362g%20n%C6%B0%E1%BB%9Bc%20%202%2F200g%20So3%20v%C3%A0o%201%20l%C3%ADt%20dung%20d%E1%BB%8Bch%20H2SO4%2017%25%20%28D%20%3D%201%2C12%20G%2FML%29

1

\(C\%_{H_2SO_4}=\dfrac{8.100\%}{92}=8,7\%\)

2

\(C\%_{H_2SO_4}=\dfrac{8.100\%}{8+92}=8\%\)

3

\(C\%_{BaCl_2}=\dfrac{15.100\%}{15+45}=25\%\)

\(a,b,m_{dd}=40+200=240\left(g\right)\\ C\%_{C_{12}H_{22}O_{11}}=\dfrac{40}{240}.100\%=16,67\%\)

nCuSO4=40/160=0,25 mol

CM CuSO4 =0,25/0,1=2,5M

nNaCl = 30/58,5=20/39 mol

nH2O = 170 /18=85/9 mol

2NaCl + 2H2O -->  Cl2 +     H2      +           2NaOH

20/39                    10/39      10/39                 20/39              mol        

 ta thấy nNaCl/2<nH2O/2

=> NaCl hết , H2O dư

=>mNaOH=20/39*20\(\approx\)20,51 g

m dd sau = 30 + 170 - 10/39*35,5-10,39*2\(\approx\)190,38 g

C% NaOh = 20,51*100/190,38=10,77%

\(a.\)

\(m_{dd}=10+40=50\left(g\right)\)

\(C\%=\dfrac{10}{50}\cdot100\%=20\%\)

\(b.\)

\(m_{KOH}=0.25\cdot56=14\left(g\right)\)

\(m_{dd_{KOH}}=14+36=50\left(g\right)\)

\(C\%_{KOH}=\dfrac{14}{50}\cdot100\%=28\%\)

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

\(n_{SO_3}=\dfrac{32}{80}=0,4\left(mol\right)\)

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

\(n_{H_2SO_4}=n_{SO_3}=0,4\left(mol\right)\)

\(m_{H_2SO_4}=0,4\cdot98=39,2\left(g\right)\)

\(m_{H_2SO_4\text{ trong dd 10%}}=\dfrac{200\cdot10}{100}=20\left(g\right)\)

\(\sum m_{H_2SO_4}=20+39,2=59,2\left(g\right)\)

\(m_{\text{ dd H2SO4 10%}}=200+39,2=239,2\left(g\right)\)

\(C\%_{\text{ dd mới}}=\dfrac{59,2}{239,2}\cdot100\%\approx24,75\%\)

Hiện tượng: SO3 được đưa vào dd H2SO4, SO3 tác dụng với H2O trong dd tạo ra sản phẩm là H2SO4.