Question

35.10 g of aluminum hydroxide is allowed to react with 53.94 g of sulfuric acid as follows:2Al(OH)3(s) + 3H2SO4(aq) -------->Al2(SO4)3(aq) + 6H2O(l)What is the maximum mass of aluminum sulfate which could be produced in the reaction?

Answer 1

62.586 gram

Step-by-step explanation:

moles of Al(OH)3 = mass / molar mass = 35.1 / (27+17x3) = 0.45 mol

moles of H2SO4 = mass / molar mass = 53.94 / (2+32+16x4) = 0.55 mol

H2SO4 is the limiting reagent (reacts completely)

⇒ moles of Al2(SO4)3 is worked out by moles of H2SO4

moles of Al2(SO4)3 = moles of H2SO4 / 3 = 0.183 mol

mass of Al2(SO4)3 = mole x molar mass = 0.183 x (27x2 + 96x3) = 62.586 gram