Question

A student 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?

Answer 1

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.