\(n_{AlCl_3}=\frac{26,7}{133,5}=0,2\left(mol\right)\)
\(PTHH:2Al+3Cl_2\underrightarrow{t^o}2AlCl_3\)
(mol)___________3_____2___
(mol)___________0,3___0,2__
\(m_{Cl_2}=71.0,3=21,3\left(g\right)\)
nAlCl3=26,7:133,5=0,3(mol)
PTHH:3Cl2+2Al-to->2AlCl3
mol:....3...............2..............2
mol:..0,45.........................0,3
mCl2=0,45.71=31,95g
nAlCl3=26,7:133,5=0,2(mol)
PTHH:3Cl2+2Al-to->2AlCl3
mol:........3....2................2
mol:.......0,3..................0,2
mCl2=0,3.71=21,3g
2Al+3Cl2---->2AlCl3
Ta có
n\(_{AlC_{ }l3}=\frac{26,7}{133,5}=0,2\left(mol\right)\)
Theo pthh
n Cl2=\(\frac{3}{2}n_{AlC_{ }l3}=0,3\left(mol\right)\)
m\(_{Cl2}=0,3.71=21,3\left(g\right)\)
PTHH : 2Al + 3Cl2 -> 2AlCl3
nAlCl3 = mAlCl3 : MAlCl3 = 26,7 : 133,5 = 0,2 ( mol )
Theo PT : 2 mol AlCl3 -> 3 mol Cl2
Bài ra : 0,2 -> x
x = 0,2 . 3 : 2 = 0,3 ( mol )
mCl2 = nCl2 . MCl2 = 0,3 . 71 = 21,3 ( g )
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