a, PTHH:    Al₂O₃ + 6HCl ➝ 2AlCl₃ + 3H₂O 

b, Áp dụng định luật bảo toàn khối lượng

=> m_HCl = m_AlCl3+ m_H2O- m_Al2O3

_⇔ m_HCl = 26,7 + 5,4 -10,2 = 21,9(g)

c, n_AlCl3= \(\dfrac{26,7}{133,5}\) = 0,2(mol)

a) PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

                \(2mol\)   \(6mol\)       \(2mol\)       \(3mol\)

                \(0,27\)      \(x\)             \(y\)             \(z\)

b) ta có: \(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{7,3}{27}=0,27\left(mol\right)\)

theo PT: \(n_{Al}=n_{AlCl_3}=0,27\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,27.133,5=36,045\left(g\right)\)

c) ta có: \(n_{H_2}=\dfrac{m_{H_2}}{M_{H_2}}=\) \(\dfrac{0,27.3}{2}=0,405\left(mol\right)\)

\(\Rightarrow V_{H_2\left(đktc\right)}=n_{H_2}.22,4=0,405.22,4=9,072\left(l\right)\)

7,3 là KL của axit mà em

thank you

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

a) 2Al + 6HCl --> 2AlCl₃ + 3H₂

b) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

2Al + 6HCl --> 2AlCl₃ + 3H₂

\(\dfrac{0,4}{3}\)<--------------------0,2

=> Al dư

\(m_{Al\left(dư\right)}=\left(0,2-\dfrac{0,4}{3}\right).27=1,8\left(g\right)\)

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

b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

\(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)

c, \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\Rightarrow m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)

d, \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

Có: \(\dfrac{0,2}{1}>\dfrac{0,45}{3}\) → Fe₂O₃ dư.

\(n_{Fe}=\dfrac{2}{3}n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)

2Al+6HCl-->2AlCl₃+3H₂

0,3----0,9---------0,3------0,45

=>n _Al=8,1\17=0,3 mol

=>V_H2=0,45.22,4=10,08l

=>m _HCl=0,9.26,5=32,85g

=>m_AlCl3=0,3.133,5=40,05g

C2 :Bảo Toàn khối lượng 

=>m _AlCl3=40,05g

\(n_{Al}=\dfrac{21,06}{27}=0,78\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(\left(mol\right)\)     \(0,78\)                    \(0,78\) 

\(m_{AlCl_3}=133,5.0,78=104,13\left(g\right)\)

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

\(n_{Al}=\dfrac{3,375}{27}=0,125\)

\(\Rightarrow n_{H_2}=\dfrac{0,125}{2}.3=0,1875mol\)      \(\Rightarrow V_{H_2}=0,1875.22,4=4,2l\)

\(n_{AlCl_3}=n_{Al}=0,125mol\)       \(\Rightarrow m_{AlCl_3}=0,125.133,5=16,6875g\)

\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2mol\)

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

0,2                      0,2          0,3    ( mol )

\(V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72l\)

\(m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,2.133,5=26,7g\)