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

b) \(n_{Al}=\dfrac{6,75}{27}=0,25\left(mol\right)\)

=> \(n_{H_2}=0,375\left(mol\right)\)

=> \(V_{H_2}=0,375.22,4=8,4\left(l\right)\)

c) \(n_{HCl}=0,75\left(mol\right)\)

=> \(m_{HCl}=0,75.36,5=27,375\left(g\right)\)

d) \(n_{AlCl_3}=0,25\left(mol\right)\)

=> \(m_{AlCl_3}=0,25.133,5=33,375\left(g\right)\)

\(a.pthh:2Al+6HCl--->2AlCl_3+3H_2\uparrow\)

b. Ta có: \(n_{Al}=\dfrac{6,75}{56}=\dfrac{27}{224}\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}.\dfrac{27}{224}=\dfrac{81}{448}\left(mol\right)\)

\(\Rightarrow V_{H_2}=\dfrac{81}{448}.22,4=4,05\left(lít\right)\)

c. Theo PT: \(n_{HCl}=3.\dfrac{27}{224}=\dfrac{81}{224}\left(mol\right)\)

\(\Rightarrow m_{HCl}=\dfrac{81}{224}.36,5\approx13,2\left(g\right)\)

d. Theo PT: \(n_{AlCl_3}=n_{Al}=\dfrac{27}{224}\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=\dfrac{27}{224}.133,5\approx16,09\left(g\right)\)

a. PTHH:   2Al +  6HCl  \(\rightarrow\)   2 AlCl₃ +  3H₂

b. Ta có: n_Al = \(\frac{6,75}{27}\) = 0,25 mol

Theo p.trình: n_H2 = \(\frac{3}{2}\)n_Al = \(\frac{3}{2}\). 0,25= 0,375 mol

\(\Rightarrow\) V_H2 = 0,375. 22,4 = 8,4 (lít).

c. Theo p.trình: n_HCl = 3.n_Al = 3.0,5= 0,75 mol

\(\Rightarrow\) m_HCl = 0,75. 36,5 = 27,375g

d. Theo p.trình: n_AlCl3 = n_Al = 0,25 mol

\(\Rightarrow\) m_AlCl3 = 0,25.133,5= 33,375g

a) pthh: 3Al+6H->2AlCl3+3H2. b) nAl=6,75/27=0,3 mol ->nH2= 3/2nAl=0,5 -> vH2=11.2 l. c) ta co nHCl=3nAl=0,9mol -> kl HCl pư =0,9×36,5=32,9g

d) nAlCl3=nAl=0,3 mol

->ko Alcl3=0.3×(27+35,5×3)=40.1g

\(2Al+6HCl\to 2AlCl_3+3H_2\)

a) Nhôm + Axit clohidric --> Nhôm clorua + Khí hidro

b) 2Al + 6HCl --> 2AlCl₃ + 3H₂

c) Theo ĐLBTKL: m_Al + m_HCl = m_AlCl3 + m_H2

=> m_HCl = 26,7 + 0,6 - 5,4 = 21,9(g)

a) 2Al +6HCl --> 2AlCl₃ + 3H₂

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

PTHH: 2Al +6HCl --> 2AlCl₃ + 3H₂

_______0,1<-0,3<----------------0,15

=> m_Al = 0,1.27 = 2,7(g)

c) n_HCl = 0,3 (mol)