Numerical answer type, answer in g, tolerance 0.01.
Answer: 2.76 g
M(KAl(SOX4)X2⋅12HX2O)=39+27+64+128+216=474 g/mol, so n=9.48/474=0.0200 mol. Two KAl(SOX4)X2 give one KX2SOX4 and one AlX2OX3, so each is 0.0100 mol. M(KX2SOX4)=174 and M(AlX2OX3)=102. Residue =0.0100×174+0.0100×102=1.74+1.02=2.76 g.