a, PTHH : \(Cl_2+Cu\rightarrow CuCl_2\)
\(n_{Cl_2}=\dfrac{V}{22.4}=\dfrac{10,8}{22,4}=0,48\left(mol\right)\)
\(n_{CuCl_2}=\dfrac{m}{M}=\dfrac{63,9}{135}=0,47\left(mol\right)\)
PTHH : \(Cl_2+Cu\rightarrow CuCl_2\)
TheoPTHH : 1 mol → 1mol
Theo baì : 0,48 mol → 0,47 mol
Tỉ lệ : \(\dfrac{0,48}{1}>\dfrac{0,47}{1}\)
=> Cl2 dư , \(CuCl_2\) hết