Question

Oxi hóa hết m gam  kim loại X thu được 1,889 m gam X_aO_b .Hỏi X là kim loại nào ?

Answer 1

\(n_X=\dfrac{m}{M_X}\left(mol\right)\)

PTHH: 2aX + bO₂ -t^o-> 2X_aO_b

____\(\dfrac{m}{M_X}\)-------------->\(\dfrac{m}{a.M_X}\)

=> \(\dfrac{m}{a.M_X}\left(a.M_X+16b\right)=1,889m\)

=> \(M_X=9.\dfrac{2b}{a}\)

Xét \(\dfrac{2b}{a}=1=>L\)

Xét \(\dfrac{2b}{a}=2=>L\)

Xét \(\dfrac{2b}{a}=3=>M_X=27\left(Al\right)\)

Answer 2

\(m_{O_2}\text{=1 , 889 − 1 = 0 , 889 g}\)

\(=>n_{O_2}=\dfrac{0,899}{32}=0,0277mol\)

\(2ãX\rightarrow bO_2\rightarrow2X_aO_b\)

\(\rightarrow n_X=\dfrac{0,0554a}{b}\)

\(\rightarrow M_X=\dfrac{1b}{\text{0 , 0554 a}}=\dfrac{18b}{a}\)

\(\text{⇒ a = 2 ; b = 3 ; M X = 27 ( A l ) }\)