Question

What is the mass of the precipitate formed when 207.86 mL of 0.250 M barium chloride is added to 150.00 mL of a 1.750 M sulfuric acid solution?

Answer 1

moles of BaCl2 = 207.86 x 0.250 / 1000 = 0.0519

moles of silfuric acid = 150 x 1.750 / 1000 = 0.2625

BaCl2 + H2SO4   ----------------> BaSO4 + 2 HCl

1 mol        1 mol                                

0.0519      0.2625                  

here limiting reagent is BaCl2 so precipitate formed according to BaCl2.

1 mol of BaCl2 -----------------   1 mol of BaSO4

0.0519 mol        -----------------    ?? BaSO4

moles of BaSO4 = 0.0519

molar mass of BaSO4 = 233.4 g/mol

moles = mass / molar mass

0.0519 = mass / 233.4

mass of precipitate BaSO4 = 12.1 g