Question

Show the calculation for the limiting reagent when 0.853g of sodium sulfate (95% pure) react with 1.707g of barium hydroxide octahydrate (87% pure) to produce a precipitate of barium sulfate. Finally, show the theoretical yield of barium sulfate.

Answer 1

mass of sodium sulfate = 0.853*95/100 = 0.81 grams

No of mol of sodium sulfate = 0.81/142.04 = 0.0057 mol

mass of barium hydroxide octahydrate = 1.707*87/100 = 1.48 grams

No of molo of barium hydroxide octahydrate = 1.48/315.46 = 0.00469 mol

Na2SO4(aq) + Ba(OH)2(aq) ----> Baso4(s) + 2H2O(l)

from equation

1 mol Na2SO4(aq) = 1 mol Ba(OH)2(aq)

limiting reagent is barium hydroxide octahydrate

No of mol of BaSO4 produced = 0.00469 mol

mass of BaSO4 produced = 0.00469*233.43 = 1.1 grams

theoretical yield of barium sulfate = 1.1 grams