How many grams of solid barium sulfate are produced when 150mL of a 1.15M sulfuric acid

Question

is reacted with 75mL of a 3.00M barium iodide solution?

Answer 1

Write a correctly balanced equation:

H2SO4(aq) + BaI2(aq) ==> BaSO4(s) + 2HI

moles H2SO4 present = 0.15 L x 1.15 mol/L = 0.1725 moles

moles BaI2 present = 0.075 L x 3.00 mol/L = 0.225 moles

Limiting reactant = H2SO4 as it will run out before the BaI2 is used up.

moles of BaSO4 that can form = 0.1725 moles H2SO4 x 1 mole BaSO4/mole H2SO4 = 0.1725 moles BaSO4 formed

mass BaSO4 = 0.1725 moles x 233 g/mole = 40.2 g (40. g to 2 sig. figs.)