Ta có: mAl = 4 (tấn) = 4000 (kg)
\(\Rightarrow n_{Al}=\dfrac{4000}{27}\left(kmol\right)\)
Theo PT: \(n_{Al_2O_3\left(LT\right)}=\dfrac{1}{2}n_{Al}=\dfrac{2000}{27}\left(kmol\right)\)
Mà: H = 90%
\(\Rightarrow n_{Al_2O_3\left(TT\right)}=\dfrac{\dfrac{2000}{7}}{90\%}=\dfrac{20000}{243}\left(kmol\right)\)
\(\Rightarrow m_{Al_2O_3\left(TT\right)}=\dfrac{20000}{243}.102=\dfrac{680000}{81}\left(kg\right)\)
Mà: %Al2O3 = 40%
\(\Rightarrow m_{bauxite}=\dfrac{m_{Al_2O_3\left(TT\right)}}{40\%}\approx20987,65\left(kg\right)\) ∼ 20,987 (tấn)
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