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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: \(n_{H_2}=\dfrac{m_{H_2}}{M_{H_2}}=\) \(\dfrac{0,27.3}{2}=0,405\left(mol\right)\)
\(\Rightarrow V_{H_2\left(đktc\right)}=n_{H_2}.22,4=0,405.22,4=9,072\left(l\right)\)
\(a.2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.n_{Al}=\dfrac{8,1}{27}=0,3mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)
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: \(n_{AlCl_3}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}\times\dfrac{67}{112}=\dfrac{67}{168}\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=\dfrac{67}{168}\times133,5=53,24\left(g\right)\)
\(a,2Al+6HCl\to 2AlCl_3+3H_2\\ b,2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ c,2Al+3Cl_2\xrightarrow{t^o}2AlCl_3\\ d,Na+H_2O\to NaOH+\dfrac{1}{2}H_2\)
2Al + 6HCl → 2AlCl3 + 3H2
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}\times0,3=0,2\left(mol\right)\)
\(\Rightarrow m_{Al}=0,2\times27=5,4\left(g\right)\)
Theo PT: \(n_{HCl}=2n_{H_2}=2\times0,3=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,6\times36,5=21,9\left(g\right)\)
Theo PT: \(n_{AlCl_3}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}\times0,3=0,2\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,2\times133,5=26,7\left(g\right)\)
\(2Al+6HCl-->2AlCl_3+3H_2\)
0,2____0,6_______0,2_______0,3
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
=>\(m_{Al}=0,2.27=5,4\left(g\right)\)
=>\(m_{HCl}=0,6.36,5=21,9\left(g\right)\)
\(m_{AlCl_3}=133,5.0,2=26,7\left(g\right)\)