Phản ứng xảy ra:

\(2Al+3O_2\rightarrow2Al_2O_3\)

Ta có:

\(n_{Al}=\frac{5,4}{27}=0,2\left(mol\right)\)

\(n_{O2}=\frac{3,6}{32}=0,1125\left(mol\right)\)

\(\Rightarrow n_{Al}>\frac{4}{3}n_{O2}\) nên Al dư.

\(\Rightarrow n_{Al2O3}=\frac{2}{3}n_{O2}=0,075\left(mol\right)\)

\(n_{Al\left(du\right)}=0,2-\frac{4}{3}n_{O2}=0,05\left(mol\right)\)

\(\Rightarrow m_{Al2O3}=0,075.\left(27.2+16.3\right)=7,65\left(g\right)\)

\(m_{Al\left(dư\right)}=0,05.27=1,35\left(g\right)\)

\(\Rightarrow\%m_{Al2O3}=\frac{7,65}{7,54+1,35}=85\%\Rightarrow\%m_{Al\left(dư\right)}=100\%-85\%=15\%\)

Hòa tan A vào HCl

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(\Rightarrow n_{HCl}=6n_{Al2O3}+3n_{Al}=0,075.6+0,05.3=0,6\left(mol\right)\)

\(\Rightarrow V_{HCl}=\frac{0,6}{4}=0,15\left(l\right)\)

a)\(n_{Al}=0,2\left(mol\right);n_{O_2}=0,1125\left(mol\right)\)

\(2Al+\frac{3}{2}O_2\rightarrow Al_2O_3\)

0,15__0,1125___0,075

\(\Rightarrow m_{Al\left(\text{dư }\right)}=1,35\left(g\right);m_{Al_2O_3}=7,65\left(g\right)\\ \Rightarrow\%Al=15\%\\ \%Al_2O_3=75\%\)

b) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

_0,05____0,15

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

0,075____0,45

\(\Rightarrow V_{HCl}=0,15\left(l\right)\)

Đáp án B

n_Ca = 0,28

n_HCl = 2n_H2SO4

=>0,28.2 = n_HCl + 2n_H2SO4 => n_HCl = 0,28 => n_H2SO4 = 0,14

m+21,14 = 11,2 + 0,14.96 + 0,28.35,5 => m = 13,44g => n_O = (13,44 - 21,14)/16 = 0,14

Bảo toàn e :  0,28.2 = 0,14.2 + 8n_NH4NO3 + 0,04.3 => n_NH4NO3 = 0,02

=>m_{ran khan} = 0,28.(40 + 62.2) + 0,02.80 = 47,52g

=>B

Đáp án A

Giúp mình câu b) với 

a) $n_{Al} = 0,2(mol)$

b)

$n_{H_2SO_4} = \dfrac{294.20\%}{98} = 0,6(mol)$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$

$Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O$

$\Rightarrow n_{Al_2O_3} = \dfrac{0,6 - 0,2.1,5}{3} = 0,1(mol)$
$m = 0,1.102 = 10,2(gam)$
$n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = 0,2(mol)$

$m_{dd} = 0,2.27 + 10,2 + 294 - 0,3.2 = 309(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,2.342}{309}.100\% = 22,1\%$

Đáp án C

a) \(n_{Al}=\dfrac{8,64}{27}=0,32\left(mol\right)\)

\(n_{HCl}=\dfrac{365.10\%}{36,5}=1\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

Xét tỉ lệ \(\dfrac{0,32}{2}< \dfrac{1}{6}\) => Al hết, HCl dư

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

         0,32-->0,96---->0,32--->0,48

=> \(V_{H_2}=0,48.22,4=10,752\left(l\right)\)

b) Trong Y chứa AlCl₃ và HCl dư

\(m_{AlCl_3}=0,32.133,5=42,72\left(g\right)\)

c) m_{dd sau pư} = 8,64 + 365 - 0,48.2 = 372,68 (g)

 \(\left\{{}\begin{matrix}C\%\left(AlCl_3\right)=\dfrac{42,72}{372,68}.100\%=11,463\%\\C\%\left(HCldư\right)=\dfrac{\left(1-0,96\right).36,5}{372,68}.100\%=0,392\%\end{matrix}\right.\)