Ta có: \(m_{H_2SO_4}=90.98\%=88,2\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{88,2}{98}=0,9\left(mol\right)\)
\(\Rightarrow n_{H^+}=0,9.2=1,8\left(mol\right)\)
Có: \(2H^++O^{2-}_{\left(trongoxit\right)}\rightarrow H_2O\)
\(\Rightarrow n_{O\left(trongoxit\right)}=\dfrac{1}{2}n_{H^+}=0,9\left(mol\right)\)
Mà: m oxit = mKL + mO (trong oxit) = 44 + 0,9.16 = 58,4 (g)