Sửa đề thành 12,5g nhé!
Gọi kl đó là R
\(n_{RO}=\frac{m}{M}=\frac{12,5}{M_R+16}\left(mol\right)\)
\(m_{ddH_2SO_4}=D.V=1,25.200=250\left(g\right)\rightarrow m_{H_2SO_4}=\frac{250.9,8}{100}=24,5\left(g\right)\rightarrow n_{H_2SO_4}=\frac{m}{M}=\frac{24,5}{98}0,25\left(mol\right)\)
Vì lượng axit dùng dư 25%
\(\Rightarrow n_{H_2SO_4}=0,25+0,25.25\%=0,3125\left(mol\right)\)
\(PTHH:RO+H_2SO_4\rightarrow RSO_4+H_2O\)
(mol) 1 1 1 1
Theo pt ta thấy:
\(n_{RO}=n_{H_2SO_4}\Leftrightarrow\frac{12,5}{M_R+16}=0,3125\Leftrightarrow M_R=\frac{12,5}{0,3125}-16=24\)
\(\rightarrow R:Mg\left(Magie\right)\)
