Question

[Bủh Bủh Dảk Dảk Lmao]

mn giúp e bài này với ạ, e cảm ơn ạ :')

 

Answer 1

[Edogawa Conan]

a, \(n_{CaCO_3}=\dfrac{41,2}{100}=0,412\left(mol\right)\)

PTHH: CaO + H₂O → Ca(OH)₂

Mol:      0,412                 0,412                        

PTHH: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

Mol:       0,412        0,412          0,412

\(m_{CaO}=0,412.56=23,072\left(g\right)\)

b, \(V_{CO_2}=0,412.22,4=9,2288\left(l\right)\)

Answer 2

[Edogawa Conan]

\(m_{Na_2CO_3}=100.16,96\%=16,96\left(g\right)\Rightarrow n_{Na_2CO_3}=\dfrac{16,96}{106}=0,16\left(mol\right)\)

\(m_{BaCl_2}=200.10,4\%=20,8\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)

PTHH: Na₂CO₃ + BaCl₂ → BaCO₃ + 2NaCl

Mol:      0,1             0,1                           0,2

Ta có: \(\dfrac{0,16}{1}>\dfrac{0,1}{1}\) ⇒ Na₂CO₃ dư, BaCl₂ hết

m_{dd sau pứ} = 100 + 200 = 300 (g)

\(C\%_{ddNaCl}=\dfrac{0,1.58,5.100\%}{300}=1,95\%\)

\(C\%_{ddNa_2CO_3}=\dfrac{\left(0,16-0,1\right).106.100\%}{300}=2,12\%\)