If 38.0 ml of 1.25 M sulphuric acid solution reacts with a 27.5 ml of 0.5 M solution containing BaCl2

Question

a write a balanced chemical equation 

b determine the theoretical yield of solid in reaction 

c if 2.5 g of was actually produce, determine percent yield of reaction 

Answer 1

a) H₂SO4(aq) + BaCl_2(aq) ==> BaSO_(s) + 2HCl(aq)

b) moles H2SO4 = 0.038L x 1.25 mol/L = 0.0475 moles

moles BaCl2 = 0.0275 L x 0.5 mol/L = 0.01375 moles

BaCl2 is the limiting reactant since they react in a mole ratio of 1:1

theoretical yield of BaSO4 = 0.01375 moles since mole ratio is 1:1 with BaCl2

theoretical yield (g) of BaSO4 = 0.01375 moles x 233 g/mole = 3.20 g

c) percent yield = 2.5 g/3.20 g (x100%) = 78%