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a.b.c.\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1 0,1 0,15 ( mol )
\(V_{H_2}=0,15.22,4=3,36l\)
\(m_{AlCl_3}=0,1.133,5=13,35g\)
d.\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,15 0,1 ( mol )
\(m_{Fe}=0,1.56=5,6g\)
a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
______0,2-->0,6--------------->0,3
=> mHCl = 0,6.36,5 = 21,9 (g)
b) VH2 = 0,3.22,4 = 6,72 (l)
a)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
______0,2-->0,6--------------->0,3
=> mHCl = 0,6.36,5 = 21,9 (g)
b) VH2 = 0,3.22,4 = 6,72 (l)
nAl = 2,7 : 27 = 0,1 (mol)
pthh : 2Al + 6HCl -> 2AlCl3 + 3H2
0,1 0,3 0,1 0,15
=> mAlCl3 = 0,1 . 133,5 = 13,35 (G)
=> VH2 = 0,15 . 22,4 = 3,36 (L)
pthh : CuO + H2 -t-> H2O + Cu
0,15 0,15
=> mCuO = 0,15 . 64 = 9,6 (G)
a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\end{matrix}\right.\)
a, Ta có: \(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
b, Ta có: \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
Bạn tham khảo nhé!
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{AlCl_3}=0,2mol\\n_{H_2}=0,3mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\end{matrix}\right.\)
a: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b: \(n_{AlCl_3}=\dfrac{26.7}{27+35.5\cdot3}=0.2\left(mol\right)\)
=>nAl=0,2(mol)
\(m=0.2\cdot27=5.4\left(g\right)\)
c: \(2\cdot n_{Al}=3\cdot n_{H_2}\Leftrightarrow n_{H_2}=\dfrac{2}{3}\cdot\dfrac{1}{5}=\dfrac{2}{15}\left(mol\right)\)
\(V=\dfrac{2}{15}\cdot22.4=\dfrac{224}{75}\left(lít\right)\)