Question

mọi người giúp với ạ

Answer 1

\(n_R=\dfrac{11,2}{M_R}\left(mol\right)\)

PTHH: 2R + 3Cl₂ --t^o--> 2RCl₃

          \(\dfrac{11,2}{M_R}\)->\(\dfrac{16,8}{M_R}\) -->\(\dfrac{11,2}{M_R}\)

=> \(\dfrac{11,2}{M_R}.\left(M_R+106,5\right)=32,5\)

=> M_R = 56 (g/mol)

=> R là Fe

\(n_{Cl_2}=\dfrac{16,8}{M_R}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

=> V_Cl2 = 0,3.22,4 = 6,72 (l)