\(2R+H_2SO_4\rightarrow R_2SO_4+H_2\left(1\right)\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\left(2\right)\\ n_{H_2SO_4\left(2\right)}=\dfrac{1}{2}n_{NaOH}=0,015\left(mol\right)\\ \Rightarrow n_{H_2SO_4\left(1\right)}=0,075-0,015=0,06\left(mol\right)\\ n_R=2n_{H_2SO_4\left(1\right)}=0,12\left(mol\right)\\ \Rightarrow M_R=\dfrac{4,68}{0,12}=39\left(Kali-K\right)\)
Ta có: \(n_{H_2SO_{4_{tham.gia}}}=\dfrac{250}{1000}.0,3=0,075\left(mol\right)\)
\(n_{NaOH}=\dfrac{60}{1000}.0,5=0,03\left(mol\right)\)
\(PTHH:\)
\(2R+H_2SO_4--->R_2SO_4+H_2\left(1\right)\)
\(H_2SO_4+2NaOH--->Na_2SO_4+2H_2O\left(2\right)\)
Theo PT(2): \(n_{H_2SO_{4_{dư}}}=\dfrac{1}{2}.n_{NaOH}=\dfrac{1}{2}.0,03=0,015\left(mol\right)\)
\(\Rightarrow n_{H_2SO_{4_{PỨ}}}=0,075-0,015=0,06\left(mol\right)\)
Theo PT(1): \(n_R=2.n_{H_2SO_4}=2.0,06=0,12\left(mol\right)\)
\(\Rightarrow M_R=\dfrac{4,68}{0,12}=39\left(\dfrac{g}{mol}\right)\)
Vậy R là kim loại kali (K)
