\(2M\left(\dfrac{1,38}{M}\right)+2H_2O\rightarrow2MOH+H_2\left(\dfrac{0,69}{M}\right)\)
\(n_M=\dfrac{1,38}{M}\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow\dfrac{0,69}{M}=0,1\)
\(\Leftrightarrow M=6,9\)
Vậy M là Li
Ta có M là kl hóa trị 1. ta có pt
2M + H2O \(\rightarrow\)2MOH + H2
nM= \(\dfrac{1,38}{M}\)mol
nH2[đktc]= \(\dfrac{V}{22,4}\)= \(\dfrac{2,24}{22,4}\)= 0,1 mol
Theo pthh ta có
nM=2n H2= 0,1.2= 0,2 MOL
\(\Leftrightarrow\)\(\dfrac{1,38}{M}\)= 0,2
\(\Leftrightarrow\) 0,2M=1,38
\(\Leftrightarrow\) M= \(\dfrac{1,38}{0,2}\)
\(\Leftrightarrow\) M= 6,9
Vậy M là Li