22. Sodium sulfate and barium chloride react according to the following equation

Question

22. Sodium sulfate and barium chloride react according to the following equation: Na2SO4 + BaCl2 NaCl + BaSO4

Use solubility rules to identify the physical state of each reactant and product.

Write the total ionic and net ionic equations for this reaction. Identify what type of reaction is occurring.

If 25.00 ml of 0.60 M sodium sulfate are mixed with 5.95 ml of 2.50 M barium chloride, what is the maximum amount of barium sulfate that can be obtained from this reaction?

If 3.12 g of barium sulfate is obtained from this reaction, what is the percent yield?

Explanation / Answer

Na2SO4(aq) + BaCl2(aq) 2NaCl(aq) + BaSO4(s)

2Na+2 (aq) +SO42-(aq) + Ba+2(aq)+2Cl-(aq) 2Na+(aq) +2C-l(aq) + BaSO4(s)

2Na+2 (aq) + Ba+2(aq) BaSO4(s) net ionic equation

Na2SO4(aq) + BaCl2(aq) 2NaCl(aq) + BaSO4(s)

no of moles of Na2So4 = molarity * volume in L

                                    = 0.6*0.025 = 0.015 moles

no of moles of BaCl2     = molarity * volume in L

                                       = 2.5*0.00595 = 0.014875 moles

from balanced equation 1 mole of Na2So4 react with 1 mole of BaCl2

BaCl2 is limiting reagent

1 mole of BaCl2 react with Na2SO4 to gives 1 mole of BaSO4

0.014875 moles of BaCl2 react with Na2So4 to gives 0.014875 moles of BaSo4

mass of BaSo4 = no of moles * gram molar mass

                         = 0.014875*233.4 = 3.47g

percentage yield = Actual yield*100/theoritical yield

                            = 3.12*100/3.47 = 89.91%

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