Question

Answer 1

ADĐLBTKL  ta có 
\(m_{O_2}=m_{MO}-m_M=7,776-6,24=1,536\left(g\right)\\ n_{O_2}=\dfrac{1,536}{32}=0,048\left(g\right)\\ pthh:2M+O_2\underrightarrow{t^o}2MO\) 
           0,096  0,048 
\(M_M=\dfrac{6,24}{0,096}=65\left(\dfrac{g}{mol}\right)\) 
=> M là Zn

Answer 2

\(2M+O_2\rightarrow\left(t^o\right)2MO\)

\(n_M=\dfrac{6,24}{M_M}\left(mol\right)\)

\(n_{MO}=\dfrac{7,776}{M_M+16}\left(mol\right)\)

Theo pt:\(n_M=n_{MO}\left(mol\right)\)

             \(\Leftrightarrow\dfrac{6,24}{M_M}=\dfrac{7,776}{M_M+16}\)

              \(\Rightarrow M_M=65\)\((g/mol)\)

`=>M` là kẽm `(Zn)`