\(m_{KClO_3\left(pư\right)}=24.5\cdot0.5=12.25\left(g\right)\)
\(n_{KClO_3}=\dfrac{12.25}{122.5}=0.1\left(mol\right)\)
\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)
\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
\(0.1...........................0.15\)
\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)
\(4............3\)
\(0.1..........0.15\)
\(LTL:\dfrac{0.1}{4}< \dfrac{0.15}{3}\Rightarrow O_2dư\)
\(m_{Al_2O_3}=0.05\cdot102=5.1\left(g\right)\)
Đúng 0
Bình luận (0)
