n_H2 = 0,75 mol

2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

0,5<---1,5<----------0,5<--0,75

\(\Rightarrow\) m_Al = 0,5.27 = 13,5 (g)

\(\Rightarrow\) V_HCl = \(\dfrac{1,5}{3}\) = 0,5 (l)

\(\Rightarrow\) m_AlCl3 = 0,5.133,5 = 66,75 (g)

\(n_{HCl}=\dfrac{150.7,3\%}{36,5}=0,3\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\\ a,n_{Al}=n_{AlCl_3}=\dfrac{2}{6}.0,3=0,1\left(mol\right)\\ \Rightarrow m_{Al}=0,1.27=2,7\left(g\right)\\ b,m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\ c,n_{H_2}=\dfrac{3}{6}.0,3=0,15\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)

nHCl=150.7,3%36,5=0,3(mol)2Al+6HCl→2AlCl3+3H2↑a,nAl=nAlCl3=26.0,3=0,1(mol)⇒mAl=0,1.27=2,7(g)b,mAlCl3=0,1.133,5=13,35(g)c,nH2=

n Al=0,25 mol

2Al + 6HCl → 2AlCl₃ + 3H₂

0,25---0,75------0,25------0,375 mol

=>VH2=0,375.22,4=8,4l

=>m AlCl3=0,25.133,5=33,375g

=>CM HCl=\(\dfrac{0,75}{2}\)=0,375M

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

b. số mol của 16,8 gam Fe:

\(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

Khối lượng của HCl:

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

c.Thể tích khí Hiđro (đktc):
\(V_{H_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)

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

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

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

c, \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\Rightarrow m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

a,\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right);n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:    0,15     0,3

Ta có: \(\dfrac{0,2}{1}>\dfrac{0,15}{1}\) ⇒ H₂ pứ hết,Fe dư

\(V_{H_2}=3,36\left(l\right)\) (đề cho)

b, ko tính đc k/lg dd ,chỉ tính đc thể tích dd

\(V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)

a) Fe + 2HCl --> FeCl₂ + H₂

b) m_tăng = m_Fe - m_H2

=> m_H2 = 8,4 - 8,1 = 0,3 (mol)

=> \(n_{H_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\)

=> V_H2 = 0,15.22,4 = 3,36 (l)

c) n_HCl(pư) = 0,3 (mol)