Bài 40: Dung dịch

\(Đặt:m_{SO_3}=a\left(g\right)\left(a>0\right)\\ SO_3+H_2O\rightarrow H_2SO_4\\ m_{H_2SO_4\left(pt\right)}=\dfrac{98}{80}a=\dfrac{49}{40}a\left(g\right)\\ m_{H_2SO_4\left(24,5\%\right)}=24,5\%.600=147\left(g\right)\\ Vì:C\%_{ddH_2SO_4}=49\%\\ \Leftrightarrow\dfrac{\dfrac{49}{40}a+147}{a+600}.100\%=49\%\\ \Leftrightarrow a=200\left(g\right)\)

Vậy cần 200 gam SO₃ hoà tan với 600 gam dd H₂SO₄ 24,5% để được dd H₂SO₄ 49%

\(Đặt:m_{ddHNO_3\left(10\%\right)}=a\left(g\right)\left(a>0\right)\\ m_{ddHNO_3\left(20\%\right)}=a+50.1,25=62,5+a\left(g\right)\\ m_{HNO_3\left(tổng\right)}=62,5.40\%+0,1a=25+0,1a\left(g\right)\\ C\%_{ddHNO_3\left(cuối\right)}=20\%\\ \Leftrightarrow\dfrac{25+0,1a}{62,5+a}.100\%=20\%\\ \Leftrightarrow a=125\left(g\right)\)

Vậy cần 125 gam dung dịch HNO₃ 10% để hoà tan 50ml dung dịch HNO₃ 40% (D=1,25g/ml) thì được dung dịch HNO₃ có nồng độ là 20%

\(n_{HCl}=0,5.0,4+0,7V=0,2+0,7V\left(mol\right)\\ C_{MddHCl\left(sau\right)}=0,5\left(M\right)\\ \Leftrightarrow\dfrac{0,2+0,7V}{0,5+V}=0,5\\ \Leftrightarrow V=0,25=250\left(ml\right)\)

a.

\(m_{ct}=m_{Na_2SO_4}+m_{NaCl}=0.2\cdot1.4\cdot142+0.2\cdot1.2\cdot58.5=53.8\left(g\right)\)

b.

\(m_{ct}=m_{HCl}+m_{NaCl}=500\cdot1.12\cdot5\%+500\cdot1.12\cdot6\%=61.6\left(g\right)\)

Gọi CTHH của muối cacbonat kim loại R là $R_2(CO_3)_n$

$MgCO_3 + 2HCl \to MgCl_2 + CO_2 + H_2O$
$R_2(CO_3)_n + 2nHCl \to 2RCl_n + nCO_2 + nH_2O$
Theo PTHH : $n_{HCl} = 2n_{CO_2} = 0,3(mol)$
$m_{dd\ HCl} = \dfrac{0,3.36,5}{7,3\%} = 150(gam)$
$m_{dd\ B} = m_A + m_{dd\ HCl} - m_{CO_2} = 14,2 + 150 - 0,15.44 = 157,6(gam)$

$\Rightarrow n_{MgCl_2} = \dfrac{157,6.6,028\%}{95} = 0,1(mol)$

Theo PTHH : 

$n_{CO_2} = n.n_{R_2(CO_3)_n} + n_{MgCO_3} = 0,15$
$\Rightarrow n_{R_2(CO_3)_n} = \dfrac{0,05}{n}(mol)$

$\Rightarrow 0,1.84 + \dfrac{0,05}{n}(2R + 60n) = 14,2$
$\Rightarrow R = 28n$
Với n = 2 thì R = 56(Fe)

$\%m_{MgCO_3} = \dfrac{0,1.84}{14,2}.100\% = 59,15\%$

$\%m_{FeCO_3} = 100\% - 59,15\% = 40,85\%$

Gọi hoá trị của R là n

Ta có: \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\) (1)

            \(R_2\left(CO_3\right)_n+2nHCl\rightarrow2RCl_n+nCO_2+nH_2O\) (2)

Theo PT (1), (2): \(n_{HCl}=2n_{CO_2}=0,3\left(mol\right)\)

=> \(m_{ddHCl}=\dfrac{0,3.36,5}{7,3\%}=150\left(g\right)\)

=> \(m_{ddspư}=150+14,2-0,15.44=157,6\left(g\right)\)

=> \(m_{MgCl_2}=157,6.6,028\%=9,5\left(g\right)\)

=> \(n_{MgCl_2}=\dfrac{9,5}{95}=0,1\left(mol\right)\)

Theo PT (1): \(n_{MgCO_3}=n_{MgCl_2}=0,1\left(mol\right)\)

=> \(m_{R_2\left(CO_3\right)_n}=14,2-0,1.84=5,8\left(g\right)\)

Theo PT (1), (2): \(n_{CO_2}=n_{MgCO_3}+n.n_{R_2\left(CO_3\right)_n}\)

=> \(n_{R_2\left(CO_3\right)_n}=\dfrac{0,15-0,1}{n}=\dfrac{0,05}{n}\left(mol\right)\)

=> \(M_{R_2\left(CO_3\right)_n}=\dfrac{5,8}{\dfrac{0,05}{n}}=116n\left(g/mol\right)\)

=> \(M_R=\dfrac{116n-60n}{2}=28n\left(g/mol\right)\)

Xét n = 2 t/m => M_R = 56 (g/mol)

=> R là Fe

=> \(\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{0,1.84}{14,2}.100\%=59,15\%\\\%m_{FeCO_3}=100\%-59,15\%=40,85\%\end{matrix}\right.\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(n_{HCl}=\dfrac{500.14,6\%}{36,5}=2\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{2}{2}\), ta được HCl dư.

Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)

⇒ n_{HCl (dư)} = 2 - 0,2 = 1,8 (mol)

a, \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, \(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

c, Ta có: m _{dd sau pư} = 5,6 + 500 - 0,1.2 = 505,8 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{12,7}{505,8}.100\%\approx2,51\%\\C\%_{HCl\left(dư\right)}=\dfrac{1,8.36,5}{505,8}.100\%\approx12,99\%\end{matrix}\right.\)

Bạn tham khảo nhé!

\(m_{HCl}=\dfrac{200.7,5\%}{100\%}=15\left(g\right)\)

\(n_{HCl}=\dfrac{15}{36,5}=0,41\left(mol\right)\)

\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)

0,205       0,41          0,205      0,205   0,205 (mol)

\(m_{CaCO_3}=0,205.100=20,5\left(g\right)\)

\(m_{CaCl_2}=0,205.111=22,755\left(g\right)\)

\(m_{CO_2}=0,205.44=9,02\left(g\right)\)

\(m_{dd}=m_{CaCO_3}+m_{ddHCl}-m_{CO_2}=20,5+200-9,02=211,48\left(g\right)\)

\(C\%_{CaCl_2}=\dfrac{22,755.100}{211,48}=10,76\%\)

Đun nóng dung dịch

A

A

A