Bài 1:

\(2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{Na}=n_{Na_2O}=0,2.2=0,4\left(mol\right)\\ a.m_{Na}=0,4.23=9,2\left(g\right)\\ b.C_{MddA}=\dfrac{0,4}{0,5}=0,8\left(M\right)\\ C\%_{ddA}=\dfrac{0,4.40}{500.1,2}.100\approx2,667\%\)

Cho Ba(OH)₂ vào muối Al sẽ có 2TH sau:

TH1: kết tủa Al(OH)₃ chưa bị hòa tan

Al³⁺ + 3OH^– → Al(OH)₃↓

→ n_Al(OH)3 = n_Al3+ → n_Al(OH)3 = xn + 0,04n

TH2: kết tủa Al(OH)₃ bị hòa tan một phần

Al³⁺                 + 3OH^–                → Al(OH)₃↓

(xn + 0,04n)→ 3(xn + 0,04n)    (xn + 0,04n)

Al(OH)₃                 + OH^–           → AlO₂^– + 2H₂O

0,952 – 3(xn + 0,04n) ←0,952

→ n_Al(OH)3 = 4xn + 0,16n – 0,952

a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\); \(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,8}{6}\) => Al hết, HCl dư

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

            0,2-->0,6---->0,2----->0,3

=> V_H2 = 0,3.22,4 = 6,72 (l)

b) \(\left\{{}\begin{matrix}m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\\m_{HCl\left(dư\right)}=\left(0,8-0,6\right).36,5=7,3\left(g\right)\end{matrix}\right.\)

=> m_{chất tan} = 26,7 + 7,3 = 34 (g)

c) m_{dd sau pư} = 5,4 + 200 - 0,3.2 = 204,8 (g)

\(\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{26,7}{204,8}.100\%=13,04\%\\C\%_{HCl\left(dư\right)}=\dfrac{7,3}{204,8}.100\%=3,56\%\end{matrix}\right.\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\) 
           0,2                       0,2       0,3 
\(V_{H_2}=0,3.22,4=6,72L\\ m_{AlCl_3}=133,5.0,2=26,7g\\ m_{\text{dd}}=5,4+200-\left(0,3.2\right)=204,8g\\ C\%=\dfrac{26,7}{204,8}.100\%=13\%\)

Mg + 2HCl → MgCl₂  + H₂ (1)

Al₂O₃ + 6HCl  →  2AlCl₃  + 3H₂O (2)

nH₂ = 2,8/22,4 = 0,125 mol 

Theo tỉ lệ phản ứng (1) => nMg = nH₂ = 0,125 mol

<=> mMg = 0,125 .24 = 3 gam và mAl₂O₃ = 8,1 - 3 =5,1 gam

%mMg = \(\dfrac{3}{8,1}\).100% = 37,03% => %mAl₂O₃ = 100 - 37,03 = 62,97%

b) nAl₂O₃ = \(\dfrac{5,1}{102}\)= 0,05 mol

=> nHCl pư = 2nMg + 6nAl₂O₃ = 0,55 mol

mHCl = 0,55.36,5 = 20,075 gam

=> mdung dịch HCl 18% = \(\dfrac{20,075}{18\%}\)= 111,53 gam

\(a,n_{Fe}=\dfrac{11,2}{56}=0,2(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=0,4(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ b,n_{H_2}=0,2(mol)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48(l)\\ c,n_{FeCl_2}=0,2(mol)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%\approx 22,93\%\)

\(n_{P_2O_5}=\dfrac{99,4}{142}=0,7\left(mol\right)\)

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

0,7        2,1           1,4 

a, \(m_{H_3PO_4}=1,4.98=137,2\left(g\right)\)

\(m_{ddH_3PO_4}=99,4+500=599,4\left(g\right)\)

Kl nước trong dd A :

\(m_{H_2O}=599,4-137,2=462,2\left(g\right)\)

\(b,C\%_{H_3PO_4}=\dfrac{137,2}{599,4}.100\%\approx22,89\%\)

\(c,C_M=\dfrac{n}{V}=\dfrac{1,4}{0,5}=2,8M\)

\(a)n_{HCl}=0,2.1,5=0,3mol\\ CaO+2HCl\rightarrow CaCl_2+H_2O\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ \Rightarrow\left\{{}\begin{matrix}2n_{CaO}+2n_{CuO}=0,3\\56n_{CaO}+80n_{CuO}=10,8\end{matrix}\right.\\ \Rightarrow n_{CaO}=n_{CaCl_2}=0,05mol;n_{CuO}=n_{CuCl_2}=0,1mol\\ \%m_{CaO}=\dfrac{0,05.56}{10,8}\cdot100=25,93\%\\ \%m_{CuO}=100-25,93=74,07\%\\ b)C_{M_{CaCl_2}}=\dfrac{0,05}{0,2}=0,25M\\ C_{M_{CuCl_2}}=\dfrac{0,1}{0,2}=0,5M\)

\(CaO+2HCl\rightarrow CaCl_2+H_2O\)

x            2x            x           x 

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

y             2y            y          y 

\(\left\{{}\begin{matrix}56x+80y=10,8\\2x+2y=0,2.1,5=0,3\end{matrix}\right.\)

\(\Rightarrow x=0,05;y=0,1\)

\(a,\%m_{CaO}=0,05.56:10,8.100\%=25,93\left(\%\right)\)

\(\%m_{CuO}=100\%-25,93\%=74,07\%\)

\(b,C_{M\left(CaCl_2\right)}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)

\(C_{M\left(CuCl_2\right)}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

a) Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\) \(\Rightarrow n_{HCl}=0,2mol\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,2\cdot36,5}{10,95\%}\approx66,67\left(g\right)\)

b) Theo PTHH: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1mol\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Zn}+m_{ddHCl}-m_{H_2}=72,97\left(g\right)\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{13,6}{72,97}\cdot100\%\approx18,64\%\)