Đáp án: A

m_O(oxit) = m_oxit – m _{kim loại} =22,3-15,1=7,2 gam

n_O(oxit)=0,45 mol => n_Cl-= 2.0,45=0,9 mol

m _muối= m_{kim loại}+ m_Cl-= 15,1+0,9.35,3= 47,05 gam

a) Gọi số mol Al, Zn là 2a, a (mol)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

          2a-->1,5a---------->a

             2Zn + O₂ --t^o--> 2ZnO

            a---->0,5a------->a

=> \(102a+81a=18,3\)

=> a = 0,1 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{0,2.27+0,1.65}.100\%=45,378\%\\\%m_{Zn}=\dfrac{0,1.65}{0,2.27+0,1.65}.100\%=54,622\%\end{matrix}\right.\)

b) \(n_{O_2}=1,5a+0,5a=0,2\left(mol\right)\)

=> \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)

a) Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow24a+27b=5,1\)  (1)

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

Bảo toàn electron: \(2a+3b=0,5\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1\cdot24}{5,1}\cdot100\%\approx47,06\%\\\%m_{Al}=52,94\%\end{matrix}\right.\)

b) Bảo toàn nguyên tố: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5\cdot36,5}{7,3\%}=250\left(g\right)\)

\(\Rightarrow V_{HCl}=\dfrac{250}{1,2}\approx208,33\left(ml\right)\)

Gọi n_Mg = x 

       n_Al = y   (mol)

\(\left\{{}\begin{matrix}24x+27y=5,1\\x+1,5y=0,25\end{matrix}\right.\)

\(\rightarrow x=0,1;y=0,1\)

\(\%m_{Mg}=\dfrac{0,1.24}{5,1}.100\%\approx47,06\%\)

\(\%m_{Al}=100\%-47,06\%=52,94\%\)

\(m_{H_2SO_4}=\dfrac{\left(0,1+0,15\right).98.100}{10}=245\left(g\right)\)

Ta có: m_{KL (trong 1 phần)} = 6,48:2 = 3,24 

- Phần 1:

m _muối = m_KL + m_Cl^- \(\Rightarrow n_{Cl^-}=\dfrac{16,02-3,24}{35,5}=0,36\left(mol\right)\)

BTNT Cl: n_HCl = n_Cl^- = 0,36 (mol)

BTNT H: n_H2 = 1/2n_HCl = 0,18 (mol)

\(\Rightarrow V_{H_2}=0,18.22,4=4,032\left(l\right)\)

- Phần 2: 

BT e, có: 2n_Zn + 2n_Mg + 3n_Al = 2n_H2 = 4n_O2

⇒ n_O2 = 0,09 (mol)

⇒ m _oxide = m_KL + m_O2 = 3,24 + 0,09.32 = 6,12 (g)

Kl Oxi: m(O) = [m(0xit) - m(Kim loai)] = (22,3 - 14,3) = 8 
====> n(0) = 8/16 = 0,5(m0l) 
Theo bảo toàn nguyên tố ta có: 
+, n(H20) = n(0) = 0,5 (mol) 
+, n(HCl) = 2n(H20) = 0,5*2 = 1 (m0l) 
Theo bảo toàn khối lượng: 
m(Oxit) + m(Axit) = m(Muối) + m(Nước) 
=====> m(Muối) = m(0xit) + m(Axit) - m(H20) = 22,3 + 36,5*1 - 18*0,5 = 49,8 
 

\(a/\\3Al+2H_2SO_4 \to Al_2(SO_4)_3+3H_2\\ Zn+H_2SO_4 \to ZnSO_4+H_2\\ n_{H_2SO_4}=\frac{400.9.8\%}{98}=0,4(mol)\\ n_{Al}=a(mol)\\ n_{Zn}=b(mol) m_{hh}=27a+65b=11,9(1)\\ n_{H_2SO_4}=1,5a+b=0,4(mol)\\ (1)(2)\\ a=0,2; b=0,1\\ b/\\ \%m_{Al}=\frac{0,2.27}{11,9}.100=45,38\%\\ \%m_{Zn}=54,62\% \)

BTKL

mO2=16.6−15=1.6(g)

nO2=\(\dfrac{1,6}{32}\)=0.05(mol)

O2+4e→2O2−

0.05....0.2

2H++2e→H2

0.2......0.2

VH2SO4=\(\dfrac{0,2}{0,2}\)=1l

nH2O=nH2SO4=0.1(mol)

BTKL

mM=16.6+0.1⋅98−0.1⋅18=24.6(g)