\(n_{CuSO_4}=\dfrac{15,2}{160}=0,095mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

0,095          0,19              0,095            0,095

\(m_{rắn}=m_{Cu\left(OH\right)_2}=0,095.98=9,31g\\ V_{ddNaOH}=\dfrac{0,19}{2}=0,095l\\ b)C_{M_{Na_2SO_4}}=\dfrac{0,095}{0,04+0,095}\approx0,7M\\ c)Cu\left(OH\right)_2\xrightarrow[t^0]{}CuO+H_2O\)

0,095                 0,095

\(m_{rắn}=m_{CuO}=0,095.80=7,6g\)

Gọi: \(\left\{{}\begin{matrix}n_{FeCl_3}=x\left(mol\right)\\n_{MgCl_2}=y\left(mol\right)\end{matrix}\right.\)

PT: \(FeCl_3+3KOH\rightarrow Fe\left(OH\right)_{3\downarrow}+3KCl\)

______x_________3x_________x (mol)

\(MgCl_2+2KOH\rightarrow Mg\left(OH\right)_{2\downarrow}+2KCl\)

____y_________2y_________y (mol)

Ta có: \(n_{KOH}=0,2.2,5=0,5\left(mol\right)\)

⇒ 3x + 2y = 0,5 (1)

m _{kết tủa} = 16,5 ⇒ 107x + 58y = 16,5 (2)

Từ (1) và (2) ⇒ x = y = 0,1 (mol)

\(\Rightarrow C_{M_{FeCl_3}}=C_{M_{MgCl_2}}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)

\(FeCl_3+3KOH\rightarrow Fe\left(OH\right)_3\downarrow+3KCl\\ MgCl_2+2KOH\rightarrow Mg\left(OH\right)_2\downarrow+2KCl\)

\(n_{KOH}=0,2\cdot2,5=0,5\left(mol\right)\)

Đặt n_FeCl₃ trong 500ml X là a mol, n_MgCl₂ trong 500ml X là b mol

\(\Rightarrow\left\{{}\begin{matrix}3a+2b=0,5\\107a+58b=16,5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\Rightarrow C_MFeCl_3=\dfrac{0,1}{0,5}=0,2\left(M\right)\\ C_MMgCl_2=\dfrac{0,1}{0,5}=0,2\left(M\right)\)

a, \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_{3\downarrow}\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)

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

b, \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe\left(OH\right)_3}=0,2.107=21,4\left(g\right)\)

\(n_{KOH}=3n_{FeCl_3}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,6}{0,2}=3\left(M\right)\)

PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

            \(CuCl_2+2KOH\rightarrow2KCl+Cu\left(OH\right)_2\downarrow\)

a+b) Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

\(\Rightarrow n_{HCl}=0,2\left(mol\right)=n_{KOH}\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2\cdot36,5}{300}\cdot100\%\approx2,43\%\\C_{M_{KOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)

c) PTHH: \(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)

Theo các PTHH: \(n_{CuO\left(lý.thuyết\right)}=n_{Cu\left(OH\right)_2}=n_{Cu}=0,1\left(mol\right)\)

\(\Rightarrow n_{CuO}=0,1\cdot95\%=0,095\left(mol\right)\) \(\Rightarrow m_{CuO}=0,095\cdot80=7,6\left(g\right)\)

a) $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{HCl} =2 n_{Fe} = 0,2.2 = 0,4(mol)$
$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$

b) $n_{H_2} = n_{FeCl_2} = n_{Fe} = 0,2(mol)

Sau phản ứng, $m_{dd} = 11,2 + 200 - 0,2.2 = 210,8(gam)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{210,8}.100\% = 12,05\%$

\(a,PTHH:2KOH+CuSO_4\rightarrow K_2SO_4+Cu\left(OH\right)_2\\ ....0,32....0,16....0,16....0,16\left(mol\right)\\ b,n_{KOH}=\dfrac{56}{56}=1\left(mol\right)\\ m_{CuSO_4}=\dfrac{400\cdot6,4\%}{100\%}=25,6\left(g\right)\\ \Rightarrow n_{CuSO_4}=\dfrac{25,6}{160}=0,16\left(mol\right)\)

Vì \(\dfrac{n_{KOH}}{2}>\dfrac{n_{CuSO_4}}{1}\) nên tính số mol theo CuSO₄

\(m_{K_2SO_4}=0,16\cdot174=27,84\left(g\right)\\ m_{Cu\left(OH\right)_2}=0,16\cdot98=15,68\left(g\right)\\ \Rightarrow m_{dd_{K_2SO_4}}=5,6+400-15,68=389,92\left(g\right)\\ \Rightarrow C\%_{K_2SO_4}=\dfrac{27,84}{389,92}\cdot100\%\approx7,14\%\)

Ta có thể viết 2 phản ứng:

Fe+6HNO₃ ⇒ Fe(NO₃)₃ +3NO₂ +3H₂O

Cu+4HNO₃ ⇒ Cu(NO₃)₂ +2NO₂ +2H₂O

Có 

Quan sát các phản ứng nhận thấy:

Dung dịch A thu được chứa Fe(NO₃)₃ và Cu(NO₃)₂.

Áp dụng định luật bảo toàn khối lượng ta có:

Đáp án B.

\(n_{BaCl_2}=0,1.1=0,1\left(mol\right)\\ BaCl_2+K_2CO_3\rightarrow BaCO_3\downarrow+2KCl\\ Chất.tan.dd.sau.p.ứ:KCl\\ n_{KCl}=2.0,1=0,2\left(mol\right)\\ V_{ddsau}=V_{ddBaCl_2}+V_{ddK_2CO_3}=100+100=200\left(ml\right)=0,2\left(l\right)\\ \Rightarrow C_{MddKCl}=\dfrac{0,2}{0,2}=1\left(M\right)\)

\(m_{HCl}=100.7,3\%=7,3\left(g\right)\Rightarrow n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

PT: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{BaCl_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\)

\(\Rightarrow C\%_{BaCl_2}=\dfrac{0,1.208}{100+100}.100\%=10,4\%\)