a)

`Zn + H_2SO_4 -> ZnSO_4 + H_2`

`ZnO + H_2SO_4 -> ZnSO_4 + H_2O`

`Zn(OH)_2 + H_2SO_4 -> ZnSO_4 + 2H_2O`

`Zn + FeSO_4 -> ZnSO_4 + Fe`

`Zn + CuSO_4 -> ZnSO_4 + Cu`

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ Zn\left(OH\right)_2+H_2SO_4\rightarrow ZnSO_4+2H_2O\\ Zn+CuSO_4\rightarrow ZnSO_4+Cu\\ Zn+FeSO_4\rightarrow ZnSO_4+Fe\)

Số mol của 5 gam \(CuSO_4.5H_2O\) tách ra \(=\dfrac{5}{250}=0.02\) mol

Trong 5 gam \(CuSO_4.5H_2O\) có

\(m_{H_2O}=0.02\times18\times5=1.8\) gam

\(m_{CuSO_4}=0.02\times160=3.2\) gam

\(m_{CuSO_4}\)tách ra \(=3.2-2.75=0.45\)  gam

\(C\%_{dd}\) bão hòa \(=\dfrac{\left(0.45\times100\right)}{0.45+1.8}=20\%\)

\(m_{H_2O}\) trong dung dịch A ban đầu \(=\dfrac{\left(1.8\times100\right)}{100-20}=2.25\) gam

\(C\%_{ddA}=\dfrac{\left(0.45\times100\right)}{0.45+2.25}=16.73\%\)

\(n_C=n_{CO_2}=0,2\left(mol\right)\Rightarrow m_C=0,2.12=2,4\left(g\right)\\ n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\Rightarrow n_H=2.0,2=0,4\left(mol\right)\Rightarrow m_H=0,4.1=0,4\left(g\right)\\ ĐLBTKL:m_A=m_C+m_H+m_O\\ \Leftrightarrow6=2,4+0,4+m_O\\ \Leftrightarrow m_O=3,2\left(g\right)\Rightarrow n_O=\dfrac{3,2}{16}=0,2\left(mol\right)\\ Đặt.CTTQ.A:C_aH_bO_c\left(a,b,c:nguyên,dương\right)\\ Ta.có:a:b:c=n_C:n_H:n_O=0,2:0,4:0,2=1:2:1\\ \Rightarrow CTĐGN:\left(CH_2O\right)_t\left(t:nguyên,dương,t>1\right)\\ Ta.có:M_{\left(CH_2O\right)_t}< 66\\ \Leftrightarrow30t< 66\\ \Leftrightarrow t< 2,2\\\Rightarrow t=2\\ \Rightarrow CTPT.A:C_2H_4O_2\)

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