Question

cho 11,2g CaO tan hết vaò nước thu đc 600 g dung dịch A (D=1,29g/ml)

a)tính C%, CM dung dịch A

b)tính khối lượng các chất thu đc khi cho dung dịch với : 2,24 lít co_{2 Dktc}

4.48 lít co2 Đktc

6,72 lít Co2 Đktc

Answer 1

https://i.imgur.com/ehnMXUD.jpg

Answer 2

https://i.imgur.com/AvXDSSm.jpg

Answer 3

CaO + H₂O → Ca(OH)₂ (1)

\(n_{CaO}=\frac{11,2}{56}=0,2\left(mol\right)\)

a) Theo pT1: \(n_{Ca\left(OH\right)_2}=n_{CaO}=0,2\left(mol\right)\)

\(\Rightarrow m_{Ca\left(OH\right)_2}=0,2\times74=14,8\left(g\right)\)

\(C\%_{Ca\left(OH\right)_2}=\frac{14,8}{600}\times100\%=2,47\%\)

\(V_{ddA}=\frac{600}{1,29}=465,12\left(ml\right)=0,46512\left(l\right)\)

\(C_{M_{Ca\left(OH\right)_2}}=\frac{0,2}{0,46512}=0,43\left(M\right)\)

b) Ca(OH)₂ + CO₂ → CaCO₃↓ + H₂O (2)

+) TD với 2,24(l) CO₂

\(n_{CO_2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)

Theo pT2: \(n_{Ca\left(OH\right)_2}=n_{CO_2}\)

Theo bài: \(n_{Ca\left(OH\right)_2}=2n_{CO_2}\)

Vì \(2>1\) ⇒ Ca(OH)₂ dư

Theo PT2: \(n_{CaCO_3}=n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=0,1\times100=10\left(g\right)\)

Theo PT2: \(n_{H_2O}=n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{H_2O}=0,1\times18=1,8\left(g\right)\)

+) TD với 4,48(l) CO₂

\(n_{CO_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT2: \(n_{Ca\left(OH\right)_2}=n_{CO_2}\)

Theo bài: \(n_{Ca\left(OH\right)_2}=n_{CO_2}\)

⇒ phản ứng hết

Theo PT2: \(n_{CaCO_3}=n_{Ca\left(OH\right)_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=0,2\times100=20\left(g\right)\)

Theo PT2: \(n_{H_2O}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{H_2O}=0,2\times18=3,6\left(g\right)\)

+) TD với 6,72(l) CO₂

\(n_{CO_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)

Theo pT2: \(n_{Ca\left(OH\right)_2}=n_{CO_2}\)

Theo bài: \(n_{Ca\left(OH\right)_2}=\frac{2}{3}n_{CO_2}\)

Vì \(\frac{2}{3}< 1\) ⇒ CO₂ dư ⇒ pứ tiếp

CO₂ + CaCO₃ + H₂O → Ca(HCO₃)₂ (3)

Theo PT2: \(n_{CO_2}pư=n_{Ca\left(OH\right)_2}=0,2\left(mol\right)\)

\(\Rightarrow n_{CO_2}dư=n_{CO_2}\left(3\right)=0,3-0,2=0,1\left(mol\right)\)

Theo PT2: \(n_{CaCO_3}=n_{Ca\left(OH\right)_2}=0,2\left(mol\right)\)

Theo PT3: \(n_{CO_2}=n_{CaCO_3}\)

Theo bài: \(n_{CO_2}=\frac{1}{2}n_{CaCO_3}\)

Vì \(\frac{1}{2}< 1\) ⇒ CaCO₃

Theo PT3: \(n_{CaCO_3}pư=n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow n_{CaCO_3}dư=0,2-0,1=0,1\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}dư=0,1\times100=10\left(g\right)\)

Theo PT3: \(n_{Ca\left(HCO_3\right)_2}=n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Ca\left(HCO_3\right)_2}=0,1\times162=16,2\left(g\right)\)