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

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right);n_{ZnCl_2}=\dfrac{27,2}{136}=0,2\left(mol\right)\)

PTHH(1): 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:     0,2                                  0,3

PTHH(2): Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,2                      0,2      0,2

Ta có: \(n_{H_2\left(1\right)}=0,5-0,2=0,3\left(mol\right)\)

\(m_{hh}=0,2.27+0,2.65=18,4\left(g\right)\)

$m_{HCl}=20\%.54,75=10,95g$

$⇒n_{HCl}=\dfrac{10,95}{36,5}=0,3mol$

$PTHH :$

$2Al+6HCl\to 2AlCl_3+3H_2$

$Theo$ $pt :$

$n_{Al}=\dfrac{1}{3}.n_{HCl}=\dfrac{1}{3}.0,3=0,1mol$

$⇒m_{Al}=0,1.27=2,7g$

$n_{H_2}=n_{HCl}=0,3mol$

$⇒V_{H_2}=0,3.22,4=6,72l$

\(a,\\ 1,\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ 2,\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ \Rightarrow n_{Al}=\dfrac{0,6.2}{3}=0,4\left(mol\right)\\ n_{Zn}=n_{H_2}=0,6\left(mol\right)\\ m_{Al}=0,4.27=10,8\left(g\right)\\ m_{Zn}=65.0,6=39\left(g\right)\\ \Rightarrow m_{Al}< m_{Zn}\\ b,Đặt:n_{Al}=n_{Zn}=1\left(mol\right)\\ \Rightarrow n_{H_2\left(1\right)}=1,5.1=1,5\left(mol\right)\\ n_{H_2\left(2\right)}=n_{Zn}=1\left(mol\right)\\ Vì:1,5>1\)

=> Cùng lấy một khối lượng kim loại Al hoặc Zn cho phản ứng thì lượng H₂ sinh ra từ phản ứng có Al sẽ nhiều hơn.

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

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

xin lỗi bạn vừa nãy nhìn nhầm xíu :v

mình làm lại này:

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

theo PT:   \(2mol\)   \(6mol\)      \(2mol\)        \(3mol\)

                 \(x\)          \(0,2\)         \(y\)               \(z\)

b) \(n_{HCl}=\dfrac{m_{HCl}}{M_{HCl}}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

\(\Rightarrow n_{AlCl_3}=\dfrac{0,2.2}{6}=0,06\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,06.133,5=8,01\left(g\right)\)

c) \(n_{H_2}=\dfrac{0,2.3}{6}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,1.22,4=2,24\left(l\right)\)

\(a.2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.n_{Al}=\dfrac{8,1}{27}=0,3mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)

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

\(m_{HCl}=0,9.36,5=32,85g\\ c.V_{H_2}=0,45.24,79=11,1555l\)

a)2Al + 6HCl → 2AlCl3 + 3H2
     0,3    0,9                         0,45          
b)
nAl= \(\dfrac{8,1}{27}=0,3\)
=>mHCl= 0,9. 36,5 = 32,85
c)
=> VH2= 0,45.22,4= 10,08l

Có j k hiểu hỏi mình nhá

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

Ta có:

nH2 = 13,4 / 22,4 = 0,6 mol

=> nAl = 2/3 . nH2 = 2/3 . 0,6= 0,4 mol

-> mAl = 0,4 . 27 = 10,8 g

+) nHCl = 2 . nH2 = 2. 0,6 = 1,2 mol

=> mHCl = 1,2 . 36,5 = 43,8 g

+) nAlCl3 = 2/3 . nH2 = 2/3 . 0,6 = 0,4 mol

=> mAlCl3 = 0,4 . 133,5 = 53,4 g

Vậy...

2Al + 6HCl → 2AlCl₃ + 3H₂

\(n_{H_2}=\dfrac{13,4}{22,4}=\dfrac{67}{112}\left(mol\right)\)

a) Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}\times\dfrac{67}{112}=\dfrac{67}{168}\left(mol\right)\)

\(\Rightarrow m_{Al}=\dfrac{67}{168}\times27=10,77\left(g\right)\)

b) Theo PT: \(n_{HCl}=2n_{H_2}=2\times\dfrac{67}{112}=\dfrac{67}{56}\left(mol\right)\)

\(\Rightarrow m_{HCl}=\dfrac{67}{56}\times36,5=43,67\left(g\right)\)

c) Theo PT: \(n_{AlCl_3}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}\times\dfrac{67}{112}=\dfrac{67}{168}\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=\dfrac{67}{168}\times133,5=53,24\left(g\right)\)

\(a,2Al+6HCl\to 2AlCl_3+3H_2\\ b,2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ c,2Al+3Cl_2\xrightarrow{t^o}2AlCl_3\\ d,Na+H_2O\to NaOH+\dfrac{1}{2}H_2\)