how many grams of solid barium sulfate form when 10mL of 0.1M of Ba(NO3)2 reacts with 10 mL of 0.5M of sulphuric acid?

Answer 1

Write the balance equation:

Ba(NO3)2 + H2SO4 ==> BaSO4(s) + 2HNO3(aq) ... balanced equation

moles Ba(NO3)2 present = 10 ml x 1 L/1000 ml x 0.1 mol/L = 0.001 moles Ba(NO3)2

moles H2SO4 present = 10 ml x 1 L/1000 ml x 0.5 mol/L = 0.005 moles H2SO4

Limiting reactant = Ba(NO3)2

Theoretical yield of BaSO4 = 0.001 mol Ba(NO3)2 x 1 mol BaSO4/mol Ba(NO3) x 233g /mol BaSO4 = 0.233 g

Thus, 0.233 g BaSO4 can form = 0.2 g BaSO4 (to 1 sig. fig.)