a, \(2Al+3Cl_2\underrightarrow{t^o}2AlCl_3\)
\(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)
b, Gọi: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\) ⇒ 27x + 56y = 7,8 (1)
Theo PT: \(n_{Cl_2}=\dfrac{3}{2}n_{Al}+\dfrac{3}{2}n_{Fe}=\dfrac{3}{2}x+\dfrac{3}{2}y=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{107}{435}\left(mol\right)\\y=\dfrac{3}{145}\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=\dfrac{107}{435}.27\approx6,64\left(g\right)\\m_{Fe}=\dfrac{3}{145}.56\approx1,16\left(g\right)\end{matrix}\right.\)
Đúng 2
Bình luận (0)
