Cho nhôm phản ứng với Clo (Cl2) sinh ra 40,05 gam Nhôm clorua (AlCl3)
a) Viết PTHH xảy ra.
b) Xác đinh hệ số tỉ lệ của PT
c) Tính khối lượng nhôm ban đầu?
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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) $n_{AlCl_3} = n_{Al} = 0,2(mol)$
$\Rightarrow m_{AlCl_3} = 0,2.133,5 = 26,7(gam)$
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) 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: \(n_{H_2}=\dfrac{m_{H_2}}{M_{H_2}}=\) \(\dfrac{0,27.3}{2}=0,405\left(mol\right)\)
\(\Rightarrow V_{H_2\left(đktc\right)}=n_{H_2}.22,4=0,405.22,4=9,072\left(l\right)\)
\(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\\ V_{kk}=4^3=64\left(dm^3\right)=64\left(l\right)\\ n_{O_2}=\dfrac{64}{5.22,4}=\dfrac{4}{7}\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
LTL: \(\dfrac{0,5}{4}< \dfrac{\dfrac{4}{7}}{3}\rightarrow\)O2 dư, lá nhôm cháy hết
\(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,5=0,25\left(mol\right)\\ m_{Al_2O_3}=0,25.102=25,5\left(g\right)\)
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)
c, \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\Rightarrow m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)
d, \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Có: \(\dfrac{0,2}{1}>\dfrac{0,45}{3}\) → Fe2O3 dư.
\(n_{Fe}=\dfrac{2}{3}n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ a,2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,1(mol);n_{HCl}=0,3(mol)\\ b,m_{Al}=0,1.27=2,7(g);m_{HCl}=0,3.36,5=10,95(g)\\ m_{AlCl_3}=0,1.133,5=13,35(g)\\ c,n_{Al}=\dfrac{16,2}{27}=0,6(mol)\\ \Rightarrow n_{H_2}=1,5n_{Al}=0,9(mol)\\ \Rightarrow V_{H_2}=0,9.22,4=20,16(l)\)
\(a,PTHH:Zn+Cl_2\rightarrow ZnCl_2\)
\(b,n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\\ Theo.PTHH:n_{Cl_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ a=m_{Cl_2}=n.M=0,4.35,5=14,2\left(g\right)\)
\(b=m_{ZnCl_2}=n.M=0,2.136=27,2\left(g\right)\)
\(c,PTHH:2Al+3Cl_2\rightarrow2AlCl_3\\ Theo.PTHH:n_{Al}=\dfrac{2}{3}.n_{Cl_2}=\dfrac{2}{3}.0,2=\dfrac{2}{15}\left(mol\right)\\ m_{Al}=n.M=\dfrac{2}{15}.27=3,6\left(g\right)\)
\(2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow \text{Số nguyên tử Al : Số phân tử }HCl=2:6=1:3 \)
a) 2Al + 3Cl2 --> 2AlCl3
b) 2:3:2
c) Số mol nhôm clorua là:
nAlCl3 = mAlCl3:MAlCl3 = 40,05:133,5 = 0,3 (mol)
--> nAl = 0,3 (mol)
Khối lượng nhôm ban đầu:
mAl = nAl.MAl = 0,3.27 = 8,1 (g)
\(a,PTHH:2Al+3Cl_2\xrightarrow{t^o}2AlCl_3\\ n_{AlCl_3}=\dfrac{40,05}{133,5}=0,3(mol)\\ b,\text{Tỉ lệ: }2:3:2\\ c,n_{Al}=n_{AlCl_3}=0,3(mol)\\ \Rightarrow m_{Al}=0,3.27=8,1(g)\)