1. \(n_{O_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Theo PT: \(n_{Al}=\dfrac{4}{3}n_{O_2}=\dfrac{2}{15}\left(mol\right)\Rightarrow m_{Al}=\dfrac{2}{15}.27=3,6\left(g\right)\)
2. \(n_{KCl\left(25\%\right)}=300.25\%=75\left(g\right)\)
Gọi: m dd KCl 10% = a (g) ⇒ mKCl (10%) = 10%a (g)
\(\Rightarrow\dfrac{75+10\%a}{a+300}=0,15\Rightarrow a=600\left(g\right)\)