PT: \(X2Ox+xH2O-->2X\left(OH\right)x\)
\(m_{dd}=4,7+95,3=100\left(g\right)\)
Gọi \(m_{X\left(OH\right)x}=a\)
=>\(\frac{a}{100}.100\%=5,6\)
\(\Rightarrow\frac{a}{100}=0,056\)
\(\Rightarrow a=5,6\left(g\right)\)
\(n_{X\left(OH\right)x}=\frac{5,6}{X+17x}\left(mol\right)\)
\(n_{X2Ox}=\frac{4,7}{2X+16x}\left(mol\right)\)
Theo pthh
\(n_{X\left(OH\right)x}=2n_{X2Ox}\)
=>\(\frac{5,6}{X+17x}=\frac{9,4}{2X+16x}\)
\(\Leftrightarrow9,4X+159,8x=11,2X+89,6x\)
\(\Leftrightarrow70,2x=1,8X\)
\(\Rightarrow X=\frac{70,2x}{1,8}\)
\(x=1\Rightarrow X=39\left(K\right)\)
Vậy X là Kali
