A2(SO4)3 + 6NaOH -> 2A(OH)3 + 3Na2(SO4)
theo pthh n muối sunfat của kl A=n kết tủa/2
<=> \(\dfrac{34,2}{2A+3\cdot96}\cdot2=\dfrac{15,6}{A+3\cdot17}\)
=> A=27
=> A là Al
A2(SO4)3 + 6NaOH → 3Na2SO4 + 2A(OH)3↓
\(n_{A\left(OH\right)_3}=\dfrac{15,6}{M_A+51}\left(mol\right)\)
Theo PT: \(n_{A_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{A\left(OH\right)_3}=\dfrac{1}{2}\times\dfrac{15,6}{M_A+51}=\dfrac{15,6}{2M_A+102}\left(mol\right)\)
Ta có: \(M_{A_2\left(SO_4\right)_3}=\dfrac{m_{A_2\left(SO_4\right)_3}}{n_{A_2\left(SO_4\right)_3}}\)
\(\Leftrightarrow2M_A+288=34,2\times\dfrac{2M_A+102}{15,6}\)
\(\Leftrightarrow31,2M_A+4492,8=68,4M_A+3488,4\)
\(\Leftrightarrow1004,4=37,2M_A\)
\(\Leftrightarrow M_A=27\left(g\right)\)
Vậy A là nguyên tố kim loại nhôm Al
