Question

When an aqueous solution of 7.00 g of BaCl₂ was added to an aqueous solution of 5.25 g of K₂SO₄, a white precipitate formed. After filtering and drying the precipitate, 6.85 g of BaSO₄ powder was obtained.

BaCl_{2 (aq)} + K₂SO_4(aq) BaSO_4(s) + 2 KCl _(aq)

What is the theoretical yield of BaSO₄? SHOW ALL WORK.

What is the percent yield of BaSO₄? SHOW ALL WORK

In determining the concentration of a sulfuric acid solution, 32.63 mL of a 0.100 M NaOH was required to titrate a 25.0 mL sample of the acid.

2 NaOH _(aq) + H₂SO₄ Na₂SO_4(aq) + 2 H₂O

What is the molarity of the H₂SO₄? SHOW ALL WORK

What volume of 0.0500 M NaOH would be required to completely neutralize 15.00 mL of 0.0817 M H₂SO₄?

Answer 1

BaCl2 (aq) + K2SO4(aq) -------> BaSO4(s) + 2 KCl (aq)
no of moles of BaCl2 = W/G.M.Wt
                        = 7/208 = 0.034 moles
no of moles of K2So4 = W/G.m.Wt
                        = 5.25/174 = 0.03 moles
1 mole of BaCl2 react with 1 mole of K2So4
0.034 moles of BaCl2 react with 0.034 moles of K2So4
   K2So4 is limiting reactant
1moles of K2So4 react with BaCl2 to gives 1 mole of BaSO4
0.03 moles of K2SO4 react with BaCL2 to gives 0.03 moles of BaSO4
mass of BaSO4 = no of moles * gram molar mass
                = 0.03*233 = 6.99g of BaSO4
   Theoritical yield = 6.99g
percentage yield = actual yield*100/theoritical yield
                    = 6.85*100/6.99 =98%
2 NaOH (aq) + H2SO4 Na2SO4(aq) + 2 H2O
2 moles       1 moles
   H2So4                     NaOH
   M1 =                     M2 = 0.1M
   V1 = 25ml                V2 = 32.63ml
   n1 = 1                   n2 = 2moles
          M1V1/n1 =   M2V2/n2
               M1   = M2V2n1/V1n2
                    = 0.1*32.63*1/25*2 = 0.06526M of H2SO4

2 NaOH (aq) + H2SO4 Na2SO4(aq) + 2 H2O
2 moles       1 moles
   H2So4                     NaOH
   M1 = 0.0817M             M2 = 0.05M
   V1 = 15ml                V2 = ml
   n1 = 1                   n2 = 2moles
          M1V1/n1 =   M2V2/n2
               V2 = M1V1n2/n1M2
                   = 0.0817*15*2/1*0.05 = 49.02ml >>>answer