Gọi CTTQ của Oxit là \(A_2O_3\)
PTHH : \(A_2O_3+3H_2SO_4\rightarrow A_2\left(SO_4\right)_3+3H_2O\)
--------1 mol---------------------1mol
Ta có : \(n_{A_2O_3}=n_{A_2\left(SO_4\right)_3}\)
\(\Leftrightarrow\dfrac{m_{A_2O_3}}{M_{A_2O_3}}=\dfrac{m_{A_2\left(SO_4\right)_3}}{M_{A_2\left(SO_4\right)_3}}\)
\(\Leftrightarrow\dfrac{20,4}{2A+48}=\dfrac{68,4}{2A+288}\)
\(\Leftrightarrow20,4\left(2A+288\right)=68,4\left(2A+48\right)\)
\(\Leftrightarrow40,8A+5875,2=136,8A+3283,2\)
\(\Leftrightarrow A=\dfrac{5875,2-3283,2}{136,8-40,8}=27\left(g/mol\right)\)
Ta thấy : A là kim loại nhôm (Al)
Vậy CTHH của Oxit là \(Al_2O_3\).
b) PTHH : \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(n_{Al_2SO_3}=\dfrac{m}{M}=\dfrac{20,4}{102}=0,2\left(mol\right)\)
\(=>n_{H_2SO_4}=0,6\left(mol\right)\)
\(=>C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,6}{0,3}=2\left(M\right)\)
