Question

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Answer 1

Câu 2 : 

\(m_{ct}=\dfrac{208.10}{100}=20,8\left(g\right)\)

\(n_{BaCl2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)

a) Pt : \(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl|\)

               1             1                1             2

             0,1           0,1            0,1            0,2

b) \(n_{H2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(m_{H2SO4}=0,1.98=9,8\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{9,8.100}{8}=122,5\left(g\right)\)

c) \(n_{BaSO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

⇒ \(m_{BaSO4}=0,1.233=23,3\left(g\right)\)

d) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

⇒ \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(m_{ddspu}=208+122,5-23,3=307,2\left(g\right)\)

\(C_{HCl}=\dfrac{7,3.100}{307,2}=2,38\)⁰/₀

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