`m_{Al} = 500 \times 28,08\% = 140,4 \text{ gam}`

`n_{Al} = \frac{140,4}{27} = 5,2 \text{ mol}`

`n_{(KAl(SO)_4)_2} = n_{Al} = 5,2 \text{ mol}`

`m_{(KAl(SO)_4)_2} = 5,2 \times 258 = 1341,6 \text{ gam}`

`m_{dd  ban đầu} = \frac{1341,6}{20,61\%} = \frac{1341,6}{0,2061} \approx 6509,4614 \text{ gam}`

`m_{\text{muối khan tách  ra}} = m \times \frac{258}{474}`

`m_{dd\ Y} = m_{dd ban đầu} - m = 6509,4614 - m`

`m_{\text{chất tan trong Y}} = 1341,6 - m \times \frac{258}{474}`

`\frac{1341,6 - m \times \frac{258}{474}}{6509,4614 - m} = 15,00\% = 0,15`

`m~~926g`