\(n_{Al}=\dfrac{21.06}{27}=0.78\left(mol\right)\)

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

\(0.78..............................1.17\)

\(V_{H_2}=1.17\cdot22.4=26.208\left(l\right)\)

thank you

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

Bn giải hêt ra ik tui hum nay bị lười pls 

B là khí hi đro nha

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

B nha

a. 2Al + 6HCl -> 2AlCl₃ + 3H₂

b. n_Al = \(\dfrac{8.1}{27}=0,3\left(mol\right)\)=> \(n_{H_2}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\)

\(V_{H_2}=0,45.22,4=10,08\left(mol\right)\)

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

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, PT: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2\)

b, Ta có: \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=6n_{Al_2O_3}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)