CHƯƠNG VI: DUNG DỊCH

Ta có: \(\left\{{}\begin{matrix}n_{M_2O_3}=\dfrac{8}{2M_M+48}=\dfrac{4}{M_M+24}\left(mol\right)\\n_{H_2SO_4}=0,3.1=0,3\left(mol\right)\\n_{NaOH}=\dfrac{50.24\%}{40}=0,3\left(mol\right)\end{matrix}\right.\)

PTHH:
\(M_2O_3+3H_2SO_4\rightarrow M_2\left(SO_4\right)_3+3H_2O\)

\(\dfrac{4}{M_M+24}\rightarrow\dfrac{12}{M_M+24}\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

0,3--------->0,15

=> \(\dfrac{12}{M_M+24}+0,15=0,3\)

=> \(M_M=56\left(g/mol\right)\)

=> M là Fe

CTHH của oxit là `Fe_2O_3`

a) $n_{FeO} = \dfrac{14,4}{72} = 0,2(mol)$
$FeO  + 2HCl \to FeCl_2 + H_2O$
$n_{HCl} = 2n_{FeO} = 0,4(mol)$
$m_{dd\ HCl} = \dfrac{0,4.36,5}{20\%} = 73(gam)$

b) $m_{dd\ sau\ pư} = 14,4 + 73 = 87,4(gam)$
$n_{FeCl_2} = n_{FeO} = 0,2(mol)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{87,4}.100\% = 29,06\%$

$Ba(OH)_2 + CO_2 \to BaCO_3 + H_2O$

$n_{CO_2} = n_{BaCO_3} = \dfrac{39,4}{197} = 0,2(mol)$
Gọi CTHH oxit sắt là $R_2O_n$
$R_2O_n + nCO \xrightarrow{t^o} 2R + nCO_2$
$n_{R_2O_n} = \dfrac{0,2}{n}(mol)$
$\Rightarrow \dfrac{0,2}{n}.(2R + 16n) = 11,6$
$\Rightarrow  R = 21n$

Với $n = \dfrac{8}{3}$ thì R = 56(Fe)

Vậy oxit sắt là $Fe_3O_4$

a) $2Na + 2H_2O \to 2NaOH + H_2$
$Na_2O + H_2O \to 2NaOH$

b)

$n_{H_2} = 0,0015(mol) \Rightarrow n_{Na} = 2n_{H_2} = 0,003(mol)$

$2NaOH + CO_2 \to Na_2CO_3 + H_2O$

$n_{Na_2CO_3} = \dfrac{6,36}{106} = 0,06(mol)$

$n_{NaOH} = 2n_{Na_2CO_3} = 0,12(mol)$

$\Rightarrow n_{Na_2O} = \dfrac{0,12 -0,003}{2} = 0,0585(mol)$

$a = 0,003.23 + 0,0585.62 = 3,696(gam)$

c) $C_{M_{NaOH}} = \dfrac{0,12}{0,6} = 0,2M$

$Mg + 2HCl \to MgCl_2 + H_2$
$MgO + 2HCl \to MgCl_2 + H_2O$
$n_{Mg} = n_{H_2} = 0,1(mol)$
$n_{MgCl_2} = \dfrac{28,5}{95} = 0,3(mol)$
$\Rightarrow n_{MgO} = 0,3 - 0,1 = 0,2(mol)$
$n_{HCl} = 2n_{MgCl_2} = 0,6(mol) \Rightarrow x = \dfrac{0,6}{0,4} = 1,5M$

$x = 0,1.24 + 0,2.40 = 10,4(gam)$

\(n_{NaOH}=0,2.1=0,2\left(mol\right)\)

\(n_{HCl}=0,4.0,5=0,2\left(mol\right)\)

PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Xét tỉ lệ: \(\dfrac{0,2}{1}=\dfrac{0,2}{1}\) => Phản ứng vừa đủ

Theo PTHH: n_NaCl = n_NaOH = 0,2 (mol)

=> \(C_{M\left(NaCl\right)}=\dfrac{0,2}{0,2+0,4}=\dfrac{1}{3}M\)

1)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂.

           0,4<--0,8<-------0,4<---0,4

           Fe₂O₃ + 3CO --t^o--> 2Fe + 3CO₂

            0,2<-------------------0,4

=> \(\left\{{}\begin{matrix}m_{Fe_2O_3}=0,2.160=32\left(g\right)\\m_{MgO}=40-32=8\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{8}{40}.100\%=20\%\\\%m_{MgO}=100\%-20\%=80\%\end{matrix}\right.\)

2)

\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)

PTHH: MgO + 2HCl ---> MgCl₂ + H₂O

           0,2---->0,4-------->0,2

m_Y = 0,4.56 + 32 = 54,4 (g)

=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,4.56}{54,4}.100\%=41,18\%\\\%m_{MgO}=100\%-41,18\%=58,82\%\end{matrix}\right.\)

3)

n_HCl = 0,4 + 0,8 = 1,2 (mol)

=> m_HCl = 1,2.36,5 = 43,8 (g)

=> \(m_{dd.HCl}=\dfrac{43,8.100}{18,25}=240\left(g\right)\)

4)

m_{dd sau pư} = 40 + 240 - 0,4.2 = 279,2 (g)

=> \(\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,4.127}{279,2}.100\%=18,2\%\\C\%_{MgCl_2}=\dfrac{0,2.95}{279,2}.100\%=6,8\%\end{matrix}\right.\)

\(a.Đặt:\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Al_2O_3}=y\left(mol\right)\\n_{FeO}=z\left(mol\right)\end{matrix}\right.\\ Tacó:80x+102y+72z=47,8\left(g\right)\left(1\right)\\ \\ Phần1:\left\{{}\begin{matrix}n_{CuO}=\dfrac{x}{2}\left(mol\right)\\n_{Al_2O_3}=\dfrac{y}{2}\left(mol\right)\\n_{FeO}=\dfrac{z}{2}\left(mol\right)\end{matrix}\right.\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ FeO+2HCl\rightarrow FeCl_2+H_2O\\ n_{HCl}=\dfrac{x}{2}.2+\dfrac{y}{2}.6+\dfrac{z}{2}.2=\dfrac{400.8,2125\%}{36,5}=0,9\left(mol\right)\left(2\right)\\ Phần2:\\ CuO+H_2SO_{4\left(đ,n\right)}\rightarrow CuSO_4+H_2O\\ Al_2O_3+3H_2SO_{4\left(đ,n\right)}\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ 2FeO+4H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+SO_2+4H_2O\\ n_{FeO}=\dfrac{z}{2}=2n_{SO_2}=2.\dfrac{1,68}{22,4}=0,15\\ \Rightarrow z=0,3\left(mol\right)\left(3\right)\\ Từ\left(1\right),\left(2\right),\left(3\right)\Rightarrow\left\{{}\begin{matrix}x=0,126\\y=0,158\\z=0,3\end{matrix}\right.\\ \Rightarrow\%CuO=21,09,\%Al_2O_3=33,72\%,\%FeO=45,19\%\\ b.m_{ddB}=\dfrac{47,8}{2}+400=123,9\left(g\right)\\ n_{CuCl_2}=\dfrac{0,126}{2}=0,063\left(mol\right)\\ n_{AlCl_3}=\dfrac{0,158}{2}.2=0,158\left(mol\right)\\ n_{FeCl_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ C\%_{CuCl_2}=\dfrac{0,063.135}{123,9}.100=6,86\%\\ C\%_{AlCl_3}=\dfrac{0,158.133.5}{123,9}.100=17,02\%\\ C\%_{FeCl_2}=\dfrac{0,15.127}{123,9}.100=15,38\%\)

\(m_{Muối}=\dfrac{500.10}{100}=50\left(g\right)\)

m_{dd sau khi trộn}  = 500 + 200 = 700 (g)

\(C\%_{dd.sau.khi.trộn}=\dfrac{50}{700}.100\%=7,143\%\)