A

D

\(n_{Al}=\dfrac{3}{27}=\dfrac{1}{9}\left(mol\right)\)

\(n_{HCl}=\dfrac{7.3}{36.5}=0.2\left(mol\right)\)

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

\(TC:\)

\(\dfrac{\dfrac{1}{9}}{2}>\dfrac{0.2}{6}\Rightarrow Aldư\)

\(n_{AlCl_3}=\dfrac{0.2\cdot2}{6}=\dfrac{1}{15}\left(mol\right)\)

\(m=\dfrac{1}{15}\cdot133.5=8.9\left(g\right)\)

nAl = 8,1 /27 = 0,3mol

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

0,3--------------->0,3------> 0,45

=> VH2 = 0,45.22,4 = 10,08 (l)

mAlCl3 = 0,3. 133,5 = 40,05 (g)

Ta có: \(C_{\%_{NaOH}}=\dfrac{m_{NaOH}}{400}.100\%=4\%\)

=> \(m_{NaOH}=16\left(g\right)\)

=> \(n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)

PTHH: HCl + NaOH ---> NaCl + H₂O

Theo PT: \(n_{NaCl}=n_{NaOH}=0,4\left(mol\right)\)

=> \(m_{NaCl}=0,4.58,5=23,4\left(g\right)\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{H_2SO_4}=0,16.5=0,8\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

LTL: \(\dfrac{0,2}{2}< \dfrac{0,8}{3}\rightarrow\)H₂SO₄ dư

Theo pt: \(\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\\V_{H_2}=0,3.22,4=6,72\left(l\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,2}{5}=0,04M\\C_{M\left(H_2SO_4.dư\right)}=\dfrac{0,8-0,3}{5}=0,1M\end{matrix}\right.\)

2Al  +  6HCl   →  2AlCl₃  +  3H₂

nAl = \(\dfrac{3,375}{27}\)= 0,125 mol

a) Theo tỉ lệ phản ứng => nH₂ = \(\dfrac{3}{2}\)nAl = 0,1875 mol

<=> V H₂ = 0,1875.22,4 = 4,2 lít

b) nAlCl₃ = nAl = 0,125 mol

=> mAlCl₃ = 0,125 . 133,5 = 16,6875 gam

 2Al + 6HCl → 2AlCl₃ + 3H₂

  2         6            2          3    

0,3       0,9         0,3       0,45

a).        n_Al= \(\dfrac{8,1}{27}\)=0,3(mol)

     ⇒  n_HCl= \(\dfrac{0,3.3}{6}\)= 0,9(mol).

     ⇒ m_HCl=n.M= 0,9 . 36.5 =32,85(g).

b).  n_AlCl3= \(\dfrac{0,9.2}{6}\)= 0,3(mol).

   ⇒m_AlCl3= n.M = 0,3 . 133,5 =40,05(g).

c). n_H2= \(\dfrac{0,3.3}{2}\)= 0,45(mol).

  ⇒V_H2= n . 22,4 = 0,45 . 22,4= 10,08(g).

Tham khảo

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

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

Theo PT: \(\left\{{}\begin{matrix}n_{HCl}=3n_{Al}=0,6\left(mol\right)\\n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\end{matrix}\right.\)

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

\(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

c, Cách 1:

Theo PT: \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

Cách 2:

Ta có: \(m_{H_2}=0,3.2=0,6\left(g\right)\)

Theo ĐLBT KL, có: m_Al + m_HCl = m_AlCl3 + m_H2

⇒ m_AlCl3 = m_Al + m_HCl - m_H2 = 5,4 + 21,9 - 0,6 = 26,7 (g)

Bạn tham khảo nhé!