Bài 13. Luyện tập chương I: Các loại hợp chất vô cơ

Câu 1:

a, PT: \(Fe\left(OH\right)_2+H_2SO_4\rightarrow FeSO_4+2H_2O\)

\(Cu\left(OH\right)_2+H_2SO_4\rightarrow CuSO_4+2H_2O\)

b, Gọi: \(\left\{{}\begin{matrix}n_{Fe\left(OH\right)_2}=x\left(mol\right)\\n_{Cu\left(OH\right)_2}=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow90x+98y=18,8\left(1\right)\)

Ta có: \(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Fe\left(OH\right)_2}+n_{Cu\left(OH\right)_2}=x+y=0,2\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe\left(OH\right)_2}=0,1.90=9\left(g\right)\\m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\end{matrix}\right.\)

\(a,2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ b,n_{H_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\Rightarrow V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c,n_{HCl}=\dfrac{6}{2}.0,2=0,6\left(mol\right)\\ m_{ddHCl}=\dfrac{0,6.36,5.100}{8,76}=250\left(g\right)\\ d,n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\Rightarrow m_{AlCl_3}=133,5.0,2=26,7\left(g\right)\\ m_{ddsau}=5,4+250-0,3.2=254,8\left(g\right)\\ C\%_{ddAlCl_3}=\dfrac{26,7}{254,8}.100\approx10,479\%\)

\(n_{Al}=\dfrac{5,4}{27}=0,2mol\\ a.2Al+6HCl->2AlCl_3+3H_2\\ b.n_{H_2}=\dfrac{3}{2}\cdot0,2=0,3mol\\ V_{H_2}=0,3\cdot22,4=6,72\left(L\right)\\ c.n_{HCl}=3n_{Al}=0,6mol\\ V_{ddHCl}=\dfrac{0,6\cdot36,5\cdot100}{8,76}=250\left(g\right)\\ d.n_{AlCl_3}=n_{Al}=0,2mol\\ m_{muối}=0,2\cdot133,5=26,7\left(g\right)\\ e.m_{ddspu}=5,4+250-0,3\cdot2=254,8\left(g\right)\\ C_{\%\left(AlCl_3\right)}=\dfrac{26,7}{254,8}\cdot100\%=10,48\%\)

a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

b) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{H_2}=\dfrac{0,2.3}{2}=0,3\left(mol\right)\)

\(V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)

c) \(n_{HCl}=\dfrac{0,2.6}{2}=0,6\left(mol\right)\)

\(m_{HCl}=0,6.36,5=21,9\left(g\right)\)

\(m_{ddHCl}=\dfrac{m_{HCl}.100\%}{C\%}=\dfrac{21,9.100\%}{8,76\%}=250\left(g\right)\)

d) \(n_{AlCl_3}=\dfrac{0,2.2}{2}=0,2\left(mol\right)\)

\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

e) \(m_{ddsau}=m_{ddHCl}+m_{Al}-m_{H_2}=250+5,4-0,3.2=254,8\left(g\right)\)

\(C\%_{AlCl_3}=\dfrac{m_{AlCl_3}}{m_{ddsau}}.100\%=\dfrac{26,7}{254,8}.100\%\approx10,48\%\)

a) $CuSO_4 + 2NaOH \to Cu(OH)_2 + Na_2SO_4$

$n_{NaOH} = \dfrac{200.6\%}{40} =0,3(mol)$
Theo PTHH, $n_{CuSO_4} = \dfrac{1}{2}n_{NaOH} = 0,15(mol)$

$\Rightarrow m_{dd\ CuSO_4} = \dfrac{0,15.160}{16\%} = 150(gam)$

b) $n_{Cu(OH)_2} = n_{CuSO_4} =0,15(mol)$

$m_{dd\ sau\ pư} = m_{dd\ NaOH} + m_{dd\ CuSO_4} - m_{Cu(OH)_2}$

$= 200 + 150 - 0,15.98 = 335,3(gam)$

$n_{Na_2SO_4} = 0,15(mol)$

$\Rightarrow C\%_{Na_2SO_4} = \dfrac{0,15.142}{335,3}.100\% = 6,35\%$

13)

$BaCO_3 \xrightarrow{t^o} BaO + CO_2$
$MgCO_3 \xrightarrow{t^o} MgO + CO_2$
$A : BaO,MgO,Al_2O_3$
$D : CO_2$
$BaO + H_2O \to Ba(OH)_2$
$Al_2O_3 + 2NaOH \to 2NaAlO_2 + H_2O$
$Al_2O_3 + Ba(OH)_2 \to Ba(AlO_2)_2 + H_2O$

$B : NaAlO_2,Ba(AlO_2)_2$

$C : MgO,Al_2O_3$

$CO_2 + NaAlO_2 + 2H_2O \to Al(OH)_3 + NaHCO_3$
$CO_2 + Ba(AlO_2)_2 + 4H_2O \to Ba(HCO_3)_2 + 2Al(OH)_3$

\(m_{BaCl_2}=\dfrac{20,8.20\%}{100\%}=4,16\left(g\right)\)

=> \(n_{BaCl_2}=\dfrac{4,16}{208}=0,02\left(mol\right)\)

\(m_{H_2SO_4}=\dfrac{39,2.10\%}{100\%}=3,92\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=\dfrac{3,92}{98}=0,04\left(mol\right)\)

\(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)

0,02          0,02          0,02          0,04

LTL: \(\dfrac{0,02}{1}< \dfrac{0,04}{1}\) => H₂SO₄ dư, BaCl₂  đủ

\(m_{H_2SO_{4\left(dư\right)}}=\left(0,04-0,02\right).98=1,96\left(g\right)\)

\(m_{BaSO_4}=0,02.233=4,66\left(g\right)\)

\(C\%_{ddH_2SO_4}=\dfrac{1,96.100}{20,8+39,2}=3,27\%\)

\(C\%_{ddHCl}=\dfrac{0,04.36,5.100}{20,8+39,2}=2,43\%\)

???

Chữa gì mà chằng chịt vậy má

\(\begin{array}{l}
FeCl_x+xNaOH\to Fe(OH)_x+xNaCl\\
Theo\,PT:\,n_{FeCl_x}=n_{Fe(OH)_x}\\
\to \dfrac{6,5}{56+35,5x}=\dfrac{4,28}{56+17x}\\
\to x=3\\
\to CTHH:\,FeCl_3
\end{array}\)