Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 2Al + 3Cl2 --to--> 2AlCl3
b) \(n_{AlCl_3}=\dfrac{13,35}{133,5}=0,1\left(mol\right)\)
PTHH: 2Al + 3Cl2 --to--> 2AlCl3
0,1<-0,15<---------0,1
=> VCl2 = 0,15.22,4 = 3,36(l)
c) mAl = 0,1.27 = 2,7(g)
\(a.PTHH:2Al+3Cl_2\overset{t^o}{--->}2AlCl_3\)
b. Ta có: \(n_{AlCl_3}=\dfrac{13,35}{133,5}=0,1\left(mol\right)\)
Theo PT: \(n_{Cl_2}=\dfrac{3}{2}.n_{AlCl_3}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\)
\(\Rightarrow V_{Cl_2}=0,15.22,4=3,36\left(lít\right)\)
c. Theo PT: \(n_{Al}=n_{AlCl_3}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al}=0,1.27=2,7\left(g\right)\)
a) $2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3$
b) $n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
Theo PTHH : $n_{Cl_2} = \dfrac{3}{2}n_{Al} = 0,3(mol)$
$\Rightarrow V_{Cl_2} = 0,3.24,79 = 7,437(lít)$
c) $n_{AlCl_3} = n_{Al} = 0,2(mol)$
$\Rightarrow m_{AlCl_3} = 0,2.133,5 = 26,7(gam)$
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 3Cl2 --to--> 2AlCl3
_____0,1-->0,15
=> VCl2 = 0,15.22,4 = 3,36(l)
\(n_{Al}=\frac{1,08}{27}=0,04mol\)
\(2Al+3Cl_2\rightarrow^{t^o}2AlCl_3\)
a) \(n_{Cl_2}=\frac{3}{2}.0,04=0,06mol\)
\(V_{Cl_2}=0,06.22,4=1,344l\)
b) Cách 1: \(m_{AlCl_3}=m_{Al}+m_{Cl_2}=1,08+71.0,06=5,34g\)
Cách 2: \(n_{AlCl_3}=n_{Al}=0,04mol\)
\(m_{AlCl_3}=0,04.133,5=5,34g\)
\(n_{Al}=\dfrac{1,08}{27}=0,04\left(mol\right)\\ 2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ a,n_{Cl_2}=\dfrac{3}{2}.0,04=0,06\left(mol\right)\\ V_{Cl_2\left(\text{Đ}KTC\right)}=0,06.22,4=1,344\left(l\right)\\ b,C1:m_{AlCl_3}=m_{Al}+m_{Cl_2}=1,08+71.0,06=5,34\left(g\right)\\ C2:n_{AlCl_3}=n_{Al}=0,04\left(mol\right)\\ m_{AlCl_3}=0,04.133,5=5,34\left(g\right)\)
\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ PTHH:4Al+3O_2-^{t^o}>2Al_2O_3\)
tỉ lệ: 4 : 3 : 2
n(mol) 0,2---->0,15---->0,1
\(m_{Al_2O_3}=n\cdot M=0,1\cdot\left(27\cdot2+16\cdot3\right)=10,2\left(g\right)\\ V_{O_2\left(dktc\right)}=n\cdot22,4=0,15\cdot22,4=3,36\left(l\right)\)
câu d đề có thiếu ko ạ?
Pt: 2Al + 6HCL -------> 2AlCl3 + 3H2 (*)
a)Theo (*) : nH2 = V : 22,4 = 3.36 : 22.4 = 0.15 (mol)
=> nAl = 2/3 . nH2 = 2/3 . 0,15 = 0,1 (mol)
=> mAl = n . M = 0,1 . 27 = 2,7 (g)
b) Theo(*): nAlCl3 = 2/3 . nH2 = 2/3 .0,15 = 0,1 (mol)
=> mAlCl3 = n . M = 0,1 . 133.5 = 13,35 (g)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PT: 2Al + 3Cl2 \(\underrightarrow{t^o}\) 2AlCl3
mol 0,1 0,15 0,1
a) \(m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
b) \(V_{Cl_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)
PTHH: 2Al = 3Cl2---> 2AlCl3
a) nAl = 2,7:27=0,1 mol
theo PTHH cứ 2 mol Al tạo thành 2 mol AlCl3
0,1 mol tạo thành 0,1 mol AlCl3
mAlCl3=0,1.133,5=13,35g
b) theo PTHH cứ 2 mol Al cần 3mol Cl2
0,1 mol Al cần 0,15 mol Cl2
VCl2=0,15.22,4=3,36(l)