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a) \(PTHH:2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\)
b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{Al}=\dfrac{0,15.2}{3}=0,1\left(mol\right)\)
\(m_{Al}=0,1.27=2,7\left(g\right)\)
c)\(n_{AlCl_3}=\dfrac{0,15.2}{3}=0,1\left(mol\right)\)
\(m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
nH2 = VH2 : 22,4 = 3,36 : 22,4 = 0,15 mol
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2
Tỉ lệ: 2 3
Pứ: ? mol 0,15
Từ pthh ta có nAl = 2/3 nH2 = 2/3 . 0,15 = 0,1 mol
=> mAl = nAl . MAl = 0,1 . 27 = 2,7g
2Al+3H2SO4->Al2(SO4)3+3H2
0,1----------------------0,075----0,15
n H2=0,15 mol
=>mAl=0,1.27=2,7g
=>m Al2(SO4)3=0,075.342=25,65g
a) PTHH: \(2Al+3H_2SO_2\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{Al}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\)
\(m_{Al}=0,1.27=2,7\left(g\right)\)
c) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)
a)
\(PTHH:2Al+6HCl->2AlCl_3+3H_2\)
1,3<---4<-------1,3<---------2
b)
\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{44,8}{22,4}=2\left(mol\right)\)
\(m_{AlCl_3}=n\cdot M=1,3\cdot\left(27+35,5\cdot3\right)=173,55\left(g\right)\)
\(m_{Al}=n\cdot M=1,3\cdot27=35,1\left(g\right)\)
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)
2Al+6HCl->2AlCl3+3H2
x----------------------------3\2x
Fe+2HCl->FeCl2+H2
y-------------------------y
=>\(\left\{{}\begin{matrix}27x+56y=21,1\\3\backslash2x+y=\dfrac{14,56}{22,4}\end{matrix}\right.\)
=>x=0,268 mol
y=0,247 mol
=>%m Al=\(\dfrac{0,268.27}{21,1}\).100=34,2938%
=>%m Fe=100-34,2938=65,7062
PTHH: 2Al + 6HCl\(\rightarrow\) 2AlCl3 + 3H2
a) Ta có: nH2=\(\frac{3,36}{22,5}\)=0,15 (mol)
nAl=0,15 (mol)
Vì: \(\frac{nAl}{2}\)=\(\frac{0,15}{2}\)=0,075
\(\frac{nH2}{3}\)=\(\frac{0,15}{3}\)=0,05
\(\rightarrow\)0,075>0,05 \(\rightarrow\)Tính số mol theo H2
\(\rightarrow\)nAlCl3=\(\frac{2}{3}\).nH2=\(\frac{2}{3}\).0,15=0,1 (mol)
\(\text{mAlCl3=0,1.133,5=13,35 (g)}\)
b) Vì Al dư =>Chất rắn sau phản ứng là Al dư
=>mAl dư=4,05-\(\frac{2}{3}\)
.nH2.27=4,05-2,7=1,35 (g)
a)
\(n_{H_2}=\frac{V_{H_2}}{22,4}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{Al}=\frac{m_{Al}}{M_{Al}}=\frac{4,05}{27}=0,15\left(mol\right)\)
\(PTHH:Al+2HCl\rightarrow AlCl_2+H_2\)
\(Theo\) \(PTHH,\) \(ta có:\)
\(n_{AlCl_2}=n_{Al}=n_{H_2}=0,15\left(mol\right)\)
\(m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,15.133,5=20,025\left(g\right)\)
b) Là AlCl3 đó bn