Question

cho 8,1 g Al cháy trong bình chứa 5,04 lít khí 02 (đkc) tính khối luongej các chất sau p ư

Answer 1

\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\\ pthh:4Al+5O_2\underrightarrow{t^o}2P_2O_5\) 
     \(LTL:\dfrac{0,3}{4}>\dfrac{0,225}{4}\) 

=> nhôm dư oxi hết 
theo pthh : \(n_{Al_2O_3}=\dfrac{2}{5}nO_2=0,09\left(mol\right)\) 
 \(m_{Al\left(p\text{ư}\right)}=\dfrac{4}{5}n_{O_2\left(p\text{ư}\right)}=0,18\left(mol\right)\\ \Rightarrow m_{Al\left(d\right)}=0,3-0,18=0,12\left(mol\right)\\ \Rightarrow m_{Ch\text{ất}r\text{ắn}}=m_{Al\left(d\right)}+m_{Al_2O_3}=\left(0,12.27\right)+\left(0,09.102\right)=12,42\left(g\right)\)