\(n_{Al}=\dfrac{10,8}{27}=0,4mol\\ 2Al+3S\xrightarrow[t^0]{}Al_2S_3\\ n_{Al_2S_3\left(lt\right)}=\dfrac{1}{2}n_{Al}=0,2mol\\ m_{Al_2S_3\left(lt\right)}=0,2.150=30g\\
H=\dfrac{25,5}{30}\cdot100\%=85\%\)
Đúng 1
Bình luận (0)