Kết tủa \(A g C l\): \(\frac{4 , 305}{143 , 5} = 0 , 03 \left(\right. m o l \left.\right)\)

Có: \(\left(\right. \overset{\overline}{M} + 35 , 5 \left.\right) . 0 , 03 = 1 , 915 \left(\right. g \left.\right) \Rightarrow \overset{\overline}{M} = 28 , 33\)

=> \(\left{\right. A : N a \\ B : K\)

=> \(\left{\right. x + y = 0 , 1 \\ 58 , 5 x + 74 , 5 y = 1 , 915 \Leftrightarrow \left{\right. x = 0 , 02 \\ y = 0 , 01\)

\(\Rightarrow X : \left{\right. m_{N a C l} = 1 , 17 \left(\right. g \left.\right) \\ m_{K C l} = 0 , 745 \left(\right. g \left.\right)\)

a) 1s2 2s2 2p6 3s2 3p1

b) aluminium nằm ở ô 13 , chu kì 3 , nhóm IIIA