Gọi kim loại cần tìm là R

Đặt \(n_{R\left(NO_3\right)_2}=n_{RCl_2}=a\left(mol\right)\)

`=>` \(\left\{{}\begin{matrix}m_{R\left(NO_3\right)_2}=a.\left(M_R+124\right)\left(g\right)\\m_{RCl_2}=a.\left(M_R+71\right)\left(g\right)\end{matrix}\right.\)

`=>` \(m_{R\left(NO_3\right)_2}>m_{RCl_2}\Rightarrow m_{R\left(NO_3\right)_2}=3,33+1,59=4,92\left(g\right)\)

`=>` \(\dfrac{m_{R\left(NO_3\right)_2}}{m_{RCl_2}}=\dfrac{a.\left(M_R+124\right)}{a.\left(M_R+71\right)}=\dfrac{4,92}{3,33}\)

`=>` \(\dfrac{M_R+124}{M_R+71}=\dfrac{4,92}{3,33}\)

`=>` \(M_R=40\left(g/mol\right)\)

`=> R: Ca`

Gọi kim loại cần tìm là A có hoá trị n

PTHH: \(2A+2nHCl\rightarrow2ACl_n+nH_2\)

Gọi \(n_{HCl}=1\left(mol\right)\Rightarrow n_{HCl\left(pư\right)}=\left(100\%-20\%\right).1=0,8\left(mol\right)\)

`=>` \(m_{ddHCl}=\dfrac{1.36,5}{3,65\%}=1000\left(g\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_A=n_{ACl_n}=\dfrac{1}{n}n_{HCl}=\dfrac{0,8}{n}\left(mol\right)\\n_{H_2}=\dfrac{1}{2}n_{HCl}=0,4\left(mol\right)\end{matrix}\right.\)

`=>` \(m_{ddspư}=1000+\dfrac{0,8M_A}{n}-0,4.2=999,2+\dfrac{0,8M_A}{n}\left(g\right)\)

`=>` \(C\%_{ACl_n}=\dfrac{\dfrac{\dfrac{0,8}{n}.\left(M_A+35,5n\right)}{n}}{999,2+\dfrac{0,8M_A}{n}}.100\%=4,973\%\)

`=>` \(M_A=28n\left(g/mol\right)\)

`n = 2 => M_A = 56`

`=> A: Fe`

a) `n_{CaCO_3} = (50)/(100) = 0,5 (mol)`

PTHH: `CaCO_3 + 2HCl -> CaCl_2 + CO_2 + H_2O`

Theo PT: `n_{CO_2} = n_{CaCl_2} = n_{CaCO_3} = 0,5 (mol)`

`=> V = 0,5.22,4 = 11,2 (l)`

b) \(C_{M\left(CaCl_2\right)}=\dfrac{0,5}{0,5}=1M\)

c) PTHH: `2NaOH + CO_2 -> Na_2CO_3 + H_2O`

Theo PT: `n_{Na_2CO_3} = n_{CO_2} =0,5 (mol)`

`=> m_{Na_2CO_3} = 0,5.106 = 53 (g)`

a) \(n_{H_2SO_4}=\dfrac{200.4,9\%}{98}=0,1\left(mol\right)\)

PTHH:              \(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\) (1)

ban đầu              0,1           0,1

phản ứng            0,1           0,1             

sau phản ứng      0              0              0,1             0,2

`=>` \(ddA\left\{{}\begin{matrix}FeCl_2:0,12\left(mol\right)\\AlCl_3:0,1\left(mol\right)\\HCl:0,2\left(mol\right)\end{matrix}\right.\)

Ta có: \(n_{NaOH}=\dfrac{37,6}{40}=0,94\left(mol\right)\)

Giả sử NaOH dư

PTHH: 

\(HCl+NaOH\rightarrow NaCl+H_2O\) (2)

0,2---->0,2--------->0,2

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\) (3)

0,12----->0,24---------->0,12--------->0,24

\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\) (4)

0,1------>0,3---------->0,1----------->0,3

`=>` \(n_{NaOH\left(pư\right)}=0,2+0,24+0,3=0,74\left(mol\right)< 0,94\)

`=> NaOH` dư, giả ửu đúng

Tiếp tục xảy ra phản ứng: \(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\) (5)

ban đầu                                 0,1             0,2

phản ứng                               0,1--------->0,1

sau phản ứng                        0                 0,1           0,1

`=>` \(\left\{{}\begin{matrix}ddC\left\{{}\begin{matrix}NaCl:0,2+0,24+0,3=0,74\left(mol\right)\\NaOH:0,1\left(mol\right)\\NaAlO_2:0,1\left(mol\right)\end{matrix}\right.\\\downarrow B:BaSO_4,Fe\left(OH\right)_2\end{matrix}\right.\)

PTHH:
\(4Fe\left(OH\right)_2+O_2\xrightarrow[]{t^o}2Fe_2O_3+4H_2O\) (6)

0,12------------------->0,06

`=>` \(m_{c.rắn}=0,1.233+0,06.160=32,9\left(g\right)\)

b) \(m_{ddC}=200+0,12.127+0,1.133,5+0,1.208+37,6-0,1.233-0,12.90=252,89\left(g\right)\)

`=>` \(m_{H_2O\left(thêm\right)}=400-252,89=147,11\left(g\right)\)

`=>` \(\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{0,74.58,5}{400}.100\%=10,8225\%\\C\%_{NaOH}=\dfrac{0,1.40}{400}.100\%=1\%\\C\%_{NaAlO_2}=\dfrac{0,1.82}{400}.100\%=2,05\%\end{matrix}\right.\)

`a.` \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) 

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1                                   0,1     ( mol )

\(Fe_xO_y+2yHCl\rightarrow xFeCl_{\dfrac{2y}{x}}+yH_2O\)

\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)

\(\Rightarrow m_{Fe_xO_y}=6,4-5,6=0,8\left(g\right)\)

Ta có: 3,2g hh + H₂ `->` 0,1g H₂O

            \(\Rightarrow\) 6,4g hh + H₂  `->` 0,2g H₂O

\(n_{H_2O}=\dfrac{0,2}{18}=\dfrac{1}{90}\left(mol\right)\)

\(Fe_xO_y+yH_2\rightarrow\left(t^o\right)xFe+yH_2O\)

\(\Rightarrow n_{O\left(Fe_xO_y\right)}=n_{H_2O}=\dfrac{1}{90}\left(mol\right)\)

Ta có:\(m_{Fe_xO_y}=56x+16.\dfrac{1}{90}=0,8\)

\(\Leftrightarrow x=\dfrac{1}{90}\)

\(\Rightarrow x:y=\dfrac{1}{90}:\dfrac{1}{90}=1:1\)

\(\Rightarrow CTHH:FeO\)

a)$NaOH + HCl \to NaCl + H_2O$
Theo PTHH : 

$n_{NaCl} = n_{NaOH} = 0,2.1 = 0,2(mol)$
$m_{NaCl} = 0,2.58,5 = 11,7(gam)$

b) $V_{dd\ X} = 0,2 + 0,2 = 0,4(lít)$
$C_{M_{NaCl}} = \dfrac{0,2}{0,4} = 0,5M$

1

a) \(\sum n_{HCl}=\dfrac{9,125+5,475}{36,5}=0,4\left(mol\right)\)

`=>` \(C_{M\left(ddC\right)}=\dfrac{0,4}{2}=0,2M\)

b) Gọi \(\left\{{}\begin{matrix}C_{M\left(ddA\right)}=xM\\C_{M\left(ddB\right)}=yM\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}V_1=\dfrac{9,125}{36,5.x}=\dfrac{0,25}{x}\left(mol\right)\\V_2=\dfrac{5,475}{36,5.y}=\dfrac{0,15}{y}\left(mol\right)\end{matrix}\right.\)

`=>` Ta có: \(\left\{{}\begin{matrix}x-y=0,4\\\dfrac{0,25}{x}+\dfrac{0,15}{y}=2\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0,1M\\y=-0,3M\end{matrix}\right.\left(\text{Loại}\right)\\\left\{{}\begin{matrix}x=0,5M\\y=0,1M\end{matrix}\right.\left(\text{Chọn}\right)\end{matrix}\right.\)

nFe tổng = 0,065; nO = \(\dfrac{4,6-mFe}{16}\) = 0,06

->nH2O = 0,06
nHCl = 0,16 = 2nH2O + 2nH2=> nH2 = 0,02
=>nFe≥ 0,02
=>nFe trong oxit≤ 0,045
=>\(\dfrac{nFe\left(oxit\right)}{nO}\)≤ \(\dfrac{0,045}{0,06}\) = \(\dfrac{2}{3}\)
=> Fe2O3 là nghiệm duy nhất.

\(n_{KOH}=0,2.0,1=0,02\left(mol\right)\\ n_{CO_2}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\\ Vì:2>\dfrac{n_{KOH}}{n_{CO_2}}=\dfrac{0,02}{0,015}=\dfrac{4}{3}>1\\ \Rightarrow Sp.2.muối:Na_2CO_3,NaHCO_3\\ PTHH:2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\left(1\right)\\ NaOH+CO_2\rightarrow NaHCO_3\left(2\right)\\ Đặt:n_{CO_2\left(1\right)}=a\left(mol\right);n_{CO_2\left(2\right)}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}a+b=0,015\\2a+b=0,02\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,005\\b=0,01\end{matrix}\right.\\ \Rightarrow m_{Na_2CO_3}=0,005.106=0,53\left(g\right)\\ m_{NaHCO_3}=0,01.84=0,84\left(g\right)\)

Ủa sao ở trên anh khi KOH dưới lại ghi NaOH z?