Question

Answer 1

$n_{H_2SO_4} = \dfrac{98.10\%}{98} = 0,1(mol)$

Gọi oxit của kim loại M là $MO$
$MO + H_2SO_4 \to MSO_4 + H_2O$
$n_{MO} = n_{H_2SO_4} = 0,1(mol)$
$M_{MO} = M + 16 = \dfrac{8}{0,1} = 80 \Rightarrow M = 64(Cu)$

Answer 2

\(MO+H_2SO_4\rightarrow MSO_4+H_2O\\ n_{H_2SO_4}=\dfrac{98.10\%}{98}=0,1\left(mol\right)\\ n_{MO}=n_{H_2SO_4}=0,1\left(mol\right)\\ \Rightarrow M_{MO}=\dfrac{8}{0,1}=80\left(\dfrac{g}{mol}\right)=M_M+16\\ \Leftrightarrow M_M=64\left(\dfrac{g}{mol}\right)\left(M:Đồng\left(Cu=64\right)\right)\\ \Rightarrow Oxit:CuO\\ \Rightarrow Chọn.C\)