2M+2xHCl---->2MClx+xH2
Ta có
n\(_M=\frac{29,25}{M}\left(mol\right)\)
n\(_{Muối}=\frac{61,2}{M+35,5x}\)
Theo pthh
n\(_M=n_{MClx}\rightarrow\frac{29,25}{M}=\frac{61,2}{M+35,5x}\)
=> 61,2M=29,25M+1047,25x
=> 31,95M=1047,25x
=> \(\frac{M}{x}=\frac{1047,25}{31,95}\)=32,7
=> M=33x
Ta có
x=1==>M=32,5(loại)
x=2---->M=65(Zn)
Vậy M là Zn