m khí = 8,96:22,4=0,4 mol
gọi số mol của 3 chât rắn lần lượt x, y ,z
Ta chỉ có Al và Mg tác dụng được vs HCl sinh ra khí H2
2Al + 6HCl -> 2AlCl3 + 3H2
2........6............2................3
x.......3x.........x...................3/2x
Mg + 2HCl -> MgCl2 + H2
1..........2..........1.................1
y.........2y...........y................y
ta có hpt
27x + 24y + 2,75=10
3/2x + y =0,4
=> x=0,261.....y=0.0083
m Al = 0,261 . 27 = 7,047g
m Mg = 0,0083. 24 = 0,203g
% Al = 7,047 :10 .100% =70,47%
%Mg= 0,203:10.100% = 2,03%
%Cu= 2,75:10.100%=27,5%

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Al₂O₃ + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂O

Cu + HCl ( Khong pu)

\(\Rightarrow\) kim loai A la Cu

\(\Rightarrow n_{CuO}=\dfrac{2,75}{80}\approx0,03\left(mol\right)\)

2Cu + O₂ \(\underrightarrow{t^0}\) 2CuO

de: 0,03 \(\leftarrow\) 0,03

\(m_{Cu}=1,92g\)

2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

de: 0,1 \(\leftarrow\) 0,15

\(m_{Al}=2,7g\)

\(m_{Al_2O_3}=10-2,7-1,92=5,38g\)

\(\%m_{Cu}=19,2\%\)

\(\%m_{Al}=27\text{%}\)

\(\%m_{Al_2O_3}=100-27-19,2=53,8\%\)

\(n_{CuO}=n_{Cu}=\dfrac{4}{80}=0,05mol\\ n_{H_2}=0,4mol\\ n_{Al}=a;n_{Zn}=b\\ 27a+65b=15,1-0,05\cdot64\\ BTe^-:3a+2b=2\cdot0,4\\ a=0,2;b=0,1\\ m_{ddHCl}=\dfrac{\left(0,6+0,2\right)\cdot36,5}{0,2}=146g\)

2Al+6HCl---->2AlCl3+3H2

Al2o3+6HCl--->2AlCl3+3H2O

Cu+HCl--> không p/u

2Cu + O2---->2CuO

ncuO=2,75/80=0.034375(mol)

Cứ 2 mol Cu---à 2 mol CuO
0.034375<------0.034375
mCu=0,034375.64=2,2(g)

--->%mCu=2,2.100/10=22%

nH2=3,36/22,4=0,15(mol)

cứ 2 mol Al----->3 mol H2
          0.1<-----0.15
mAl :0,1.27=2.7(g)

--->%mAl=2,7.100/10=27%

---->%mAl2o3=100%-27%-22%=51%

  cu ko tac dung voi HCl=>2,75g la khoi luong cua Cu => %Cu = 2,75/10*100=27,5% 
n H2 = 3,36/22,4= 0,15 mol 
pt 2Al + 6HCl -----> 2AlCl3 + 3H2 
=> n Al = 2/3n H2 = 2/3 *0,15= 0,1 mol => m Al = 0,1 *27 = 2,7g 
=> % Al = 2,7/10*100= 27% 
=> %Al2O3 = 100%-27,5%-27%=45,5% 

\(n_{CuO}=\dfrac{2,75}{80}=0,034375mol\\ 2Cu+O_2\rightarrow2CuO\)

0,034375 0,0171875 0,034375

\(m_{Al,Mg}=10-0,034375.64=7,8g\\ n_{H_2}=\dfrac{9,916}{22,79}=0,4mol\\ n_{Al}=a;n_{Mg}=b\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+24b=7,8\\1,5a+b=0,4\end{matrix}\right.\\ \Rightarrow a=0,2;b=0,1\\ \%m_{Al}=\dfrac{27.0,2}{10}\cdot100=54\%\\ \%m_{Mg}=\dfrac{0,1.24}{10}\cdot1=24\%\\ \%m_{Cu}=100-54-24=22\%\\ 4Al+3O_2\xrightarrow[]{t^0}2Al_2O_3\left(2\right)\\ n_{O_2\left(2\right)}=\dfrac{0,2.3}{4}=0,15mol\\ 2Mg+O_2\xrightarrow[]{t^0}2MgO\left(3\right)\\ n_{O_2\left(3\right)}=\dfrac{0,1}{2}=0,05mol\\ V_{O_2}=\left(0,0171875+0,15+0,05\right).24,79\approx5,384l\)

Chọn đáp án D

Bài 6 : Chất rắn không tan là Cu

$m_{Cu} = 6,4(gam)$

Gọi $n_{Al} = a(mol) ; n_{Mg} = b(mol) \Rightarrow 27a + 24b + 6,4 = 14,2(1)$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 +3 H_2$
$Mg + H_2SO_4 \to MgSO_4 + H_2$
$n_{H_2} = 1,5a + b = 0,4(2)$

Từ (1)(2) suy ra a = 0,2 ; b = 0,1

$\%m_{Al} = \dfrac{0,2.27}{14,2}.100\% = 38,03\%$
$\%m_{Mg} = \dfrac{0,1.24}{14,2}.100\% =16,9\%$
$\%m_{Cu} = 100\% -38,03\% - 16,9\% = 45,07\%$

Bài 7 : 

Gọi $n_{CuO} = a(mol) ; n_{ZnO} = b(mol) \Rightarrow 80a + 81b = 12,1(1)$
$CuO + 2HCl \to CuCl_2 + H_2O$
$ZnO + 2HCl \to ZnCl_2 + H_2O$
$n_{HCl} = 2a + 2b = 0,1.3 = 0,3(2)$

Từ (1)(2) suy ra a= 0,05 ; b = 0,1

$\%m_{CuO} = \dfrac{0,05.80}{12,1}.100\% = 33,06\%$
$\%m_{ZnO}  = 100\% - 33,06\% = 66,94\%$