Question

Answer 1

\(n_{Fe}=\dfrac{13,44}{56}=0,24\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\) 
         0,24                                0,24 
\(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\\ pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\) 
\(LTL:\dfrac{0,3}{1}>\dfrac{0,24}{1}\) 
=> CuO dư 
THEO PTHH : \(n_{CuO\left(p\text{ư}\right)}=n_{Cu}=n_{H_2}=0,24\left(mol\right)\) 
\(n_{CuO\left(d\right)}=0,3-0,24=0,06\left(mol\right)\)
 \( \Rightarrow m_{KL}=\left(0,24.64\right)+\left(0,06.80\right)=20,16\left(g\right)\)

Answer 2

hỏi từng câu thôi bn