a. CTTQ : \(X_2O_3\)
\(M_A=\dfrac{20.4}{0,2}=102\left(g/mol\right)\)
\(\Rightarrow X=\dfrac{102-16.3}{2}=27\left(dvC\right)\)
\(CTHH:Al_2O_3\)
b. CTTQ : \(YO_2\)
\(n_B=\dfrac{4.48}{22,4}=0,2\left(mol\right)\)
\(M_B=\dfrac{8.8}{0,2}=44\left(g/mol\right)\)
\(\Rightarrow Y=44-16.2=12\left(dvC\right)\)
\(CTHH:CO_2\)
a)
ap dung quy tac hoa tri
=> CTHH: X2O3
\(M_{X_2O_3}=\dfrac{m}{n}=\dfrac{20,4}{0,2}=102\left(\dfrac{g}{mol}\right)\)
vi khoi luong mol co chi so trung voi nguyen tu hoac phan tu chat do
=> PTK(X2O3)=102(dvC)
=> PTK(X2)=102-16x3=54(dvC)
=> NTK(X)=54:2=27(dvC)
=> X la nhom (Al)
b)
ap dung QTHT
=> CTHH: YO2
\(n_{YO_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ M_{YO_2}=\dfrac{m}{n}=\dfrac{8,8}{0,2}=44\left(\dfrac{g}{mol}\right)\)
vi khoi luong mol co chi so trung voi nguyen tu hoac phan tu chat do
=> PTK(YO2)=44(dvC)
=> NTK(Y)=44-16x2=12(dvC)
=> Y la cacbon (C)
