Astudent performs a titration analysis for the following balanced reaction: 3 h2so4 + 2 al(oh)3 → al2(so4)3 + 6 h2o if a 200.0g sample of al(oh)3 is used to react with excess h2so4, how many grams of aluminum sulfate are produced?

The molar mass of Al(OH)3 = 27+(16+1)*3 = 78 g/mol. Dividing the 200.0 g by the molar mass of 78 g/mol = 2.564 moles. From the balanced equation, 2 moles of Al(OH)3 are equivalent to 1 mole of Al2(SO4)3, so if 2.564 moles of Al(OH3) are used, we divide by 2 to find that 1.282 moles of Al2(SO4)3 are formed. The molar mass of Al2(SO4)3 is 27*2+(32.07+16*4)*3 = 342.21 g/mol, so multiplying this by 1.282 moles gives 438.71 grams of aluminum sulfate produced.

They are all full and don't need any more electrons in there outer shell.