Bài 42: Nồng độ dung dịch

1.

Bài ra: \(m_{CuSO_4.5H_2O}=25g\)

Có: \(M_{C\text{uS}O_4.5H_2O}=160+5.18=250\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow n_{CuSO_4.5H_2O\:}=\dfrac{25}{250}=0,1\left(mol\right)\)

\(\Rightarrow n_{C\text{uS}O_4}=1.0,1=0,1\left(mol\right)\)

\(\Rightarrow m_{C\text{uS}O_4}=0,1.160=16\left(g\right)\)

\(\Rightarrow m_{H_2O}=25-16=9\left(g\right)\)

Vậy \(m_{ct}=m_{C\text{uS}O_4trong.tinh.th\text{ể}}=16\left(g\right)\)

\(m_{dm}=100+9=109\left(g\right)\)

\(m_{\text{dd}}=100+25=125\left(g\right)\)

2.(bạn ghi sai rồi Ba(OH)₂chứ ko có (BaOH)₂nhé)

Có: \(m_{Ba\left(OH\right)_2}=34,2\left(g\right)\)

\(m_{H_2O}=200g\)

\(\Rightarrow m_{\text{dd}}=34,2+200=234,2\left(g\right)\)

\(\Rightarrow C\%_{_{ }}=\dfrac{34,2}{234,2}.100\%\approx14,603\%\)

1) Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O (1)

K₂SO₃ + 2HCl → 2KCl + SO₂ + H₂O (2)

\(n_{HCl}=0,25\times4=1\left(mol\right)\)

Gọi \(x,y\) lần lượt là số mol của Na₂CO₃ và K₂SO₃

Ta có: \(\left\{{}\begin{matrix}106x+158y=58,2\\2x+2x=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,1\end{matrix}\right.\)

Vậy: \(n_{Na_2CO_3}=0,4\left(mol\right)\) và \(n_{K_2SO_3}=0,1\left(mol\right)\)

Theo PT1: \(n_{CO_2}=n_{Na_2CO_3}=0,4\left(mol\right)\)

Theo PT2: \(n_{SO_2}=n_{K_2SO_3}=0,1\left(mol\right)\)

\(\Rightarrow V_{hhC}=22,4\times\left(0,4+0,1\right)=11,2\left(l\right)\)

2) \(m_{Na_2CO_3}=0,4\times106=42,4\left(g\right)\)

\(\Rightarrow\%Na_2CO_3=\dfrac{42,4}{58,2}\times100\%=72,85\%\)

\(m_{K_2SO_3}=0,1\times158=15,8\left(g\right)\)

\(\Rightarrow\%K_2SO_3=\dfrac{15,8}{58,2}\times100\%=27,15\%\)

3) \(m_{ddHCl}=250\times1,2=300\left(g\right)\)

\(\Rightarrow m_{ddB}=58,2+300=358,2\left(g\right)\)

Theo PT1: \(n_{NaCl}=2n_{Na_2CO_3}=2\times0,4=0,8\left(mol\right)\)

\(\Rightarrow m_{NaCl}=0,8\times58,5=46,8\left(g\right)\)

Theo PT2: \(n_{KCl}=2n_{K_2SO_3}=2\times0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{KCl}=0,2\times74,5=14,9\left(g\right)\)

\(\Rightarrow m_{ct}=14,9+46,8=61,7\left(g\right)\)

\(\Rightarrow C\%_{ddB}=\dfrac{61,7}{358,2}\times100\%=17,23\%\)

4) Ta có: \(d_{\dfrac{hhC}{KK}}=\dfrac{M_{hhC}}{M_{KK}}=\dfrac{M_{CO_2}+M_{SO_2}}{M_{KK}}=\dfrac{44+64}{29}\approx4\)

Vậy hỗn hợp C nặng hơn không khí và nặng hơn gần 4 lần

theo bài ta có :

mK2O= 28,2g

=> nK2O = 0,3mol

pthh:

K2O + H2O -> 2KOH

1mol....1mol......2mol

0,3mol..0,3mol...0,6mol

=> nKOH = 0,6mol

ta có : mddA = 28,2 +