a) \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

m_Cu = m_Y = 9,6 (g)

Gọi số mol Al, Mg là a, b

=> 27a + 24b = 14,7 - 9,6 = 5,1 (g)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

            a-->3a-------->a------>1,5a 

            Mg + 2HCl --> MgCl₂ + H₂

            b--->2b------->b----->b

=> 1,5a + b = 0,25

=> a = 0,1; b = 0,1

=> \(\left\{{}\begin{matrix}m_{Al}=0,1.27=2,7\left(g\right)\\m_{Mg}=0,1.24=2,4\left(g\right)\\m_{Cu}=9,6\left(g\right)\end{matrix}\right.\)

b) n_HCl(PTHH) = 3a + 2b = 0,5 (mol)

=> n_{HCl(thực tế)} = \(\dfrac{0,5.120}{100}=0,6\left(mol\right)\)

=> \(C_{M\left(HCl\right)}=\dfrac{0,6}{0,2}=3M\)

c) \(\left\{{}\begin{matrix}C_{M\left(AlCl_3\right)}=\dfrac{0,1}{0,2}=0,5M\\C_{M\left(MgCl_2\right)}=\dfrac{0,1}{0,2}=0,5M\\C_{M\left(HCldư\right)}=\dfrac{0,6-0,5}{0,2}=0,5M\end{matrix}\right.\)

d) \(n_{Cu}=\dfrac{9,6}{64}=0,15\left(mol\right)\)

PTHH: \(Cu+Cl_2\underrightarrow{t^o}CuCl_2\)

         0,15-->0,15

=> \(V_{Cl_2}=0,15.22,4=3,36\left(l\right)\)

\(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)

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

\(\Rightarrow m_{Fe_2O_3}=16\left(g\right)\)

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

Lại có : \(n_{HCl}=2n_{H_2}+3n_{FeCl_3}=0,8\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=292\left(g\right)\)

\(\Rightarrow V=\dfrac{2920}{11}\left(ml\right)=\dfrac{73}{275}\left(l\right)\)

\(\Rightarrow C_{MFeCl_3}=\dfrac{0,2}{\dfrac{73}{275}}=\dfrac{55}{73}\left(M\right)\)

\(n_{H_2} = 1(mol)\)

Fe + 2HCl → FeCl₂ + H₂

1.........2.......................1.............(mol)

⇒ m_Fe = 1.56 = 56 > m_{hỗn hợp} = 21,6

(Sai đề)

a, Hiện tượng: có kết tủa trắng.

\(n_{KCl}=0,4\cdot2=0,8\left(mol\right);n_{AgNO_3}=2\cdot0,1=0,2\left(mol\right)\\ PTHH:KCl+AgNO_3\rightarrow KNO_3+AgCl\downarrow\\ ....0,2....0,2.....0,2....0,2\left(mol\right)\)

Vì \(\dfrac{n_{KCl}}{1}>\dfrac{n_{AgNO_3}}{1}\) nên sau phản ứng KCl dư, tính theo AgNO₃

\(\Rightarrow m_{AgCl}=0,2\cdot143,5=28,7\left(g\right)\)

\(b,V_{dd\left(sau.p/ứ\right)}=V_{KNO_3}=V_{KCl}+V_{AgNO_3}=0,4+0,1=0,5\left(l\right)\\ \Rightarrow C_{M_{KNO_3}}=\dfrac{0,2}{0,5}=0,4M\)

BTKL

m_X + m_{dd HNO3} = m_{dd X} + m_H2O + m↑

=> m_{dd X} = 11,6 + 87,5 – 30 . 0,1 – 46 . 0,15 = 89,2g

=> C%Fe(NO₃)₃ = 13,565%

a)

\(n_{H_2}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

          0,03<------------0,03<----0,015

=> \(\%m_{Na}=\dfrac{0,03.23}{1,31}.100\%=52,67\%\)

=> \(\%m_{Na_2O}=100\%-52,67\%=47,33\%\)

b)

\(n_{Na_2O}=\dfrac{1,31.47,33\%}{62}=0,01\left(mol\right)\)

PTHH: Na₂O + H₂O --> 2NaOH

            0,01----------->0,02

=> n_NaOH = 0,03 + 0,02 = 0,05 (mol)

m_{dd sau pư} = 1,31 + 18,72 - 0,015.2 = 20 (g)

=> \(C\%_{dd.NaOH}=\dfrac{0,05.40}{20}.100\%=10\%\)

\(V_{dd.NaOH}=\dfrac{20}{1,2}=\dfrac{50}{3}\left(ml\right)=\dfrac{1}{60}\left(l\right)\) 

\(C_{M\left(dd.NaOH\right)}=\dfrac{0,05}{\dfrac{1}{60}}=3M\)