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Chapter 3: Q3.107P (page 89)

How many moles of excess reactant are present when 350 mL of 0.210 M sulfuric acid reacts with 0.500 L of 0.196 M sodium hydroxide to form water and aqueous sodium sulfate?

Short Answer

Expert verified

A 2.45×10-2mol H2SO4of excess reactant are present when 350 mL of 0.210 M sulfuric acid reacts with 0.500 L of 0.196 M sodium hydroxide to form water and aqueous sodium sulfate.

Step by step solution

01

Calculating the number of moles of BaSO4

Calculate the mole number produced using the given reactants:

Moles of Na2SO4=35×10-3L H2SO4soln×0.21 mol H2SO41 L H2SO4soln×1 mol Na2SO41 mol H2SO4                                         =0.0735 mol Na2SO4\hfill                                         =50.5 l NaOHsoln×0.196 mol NaOH1 L NaOH soln×1 mol Na2SO42 mol NaOH                                         =0.049mol Na2SO4.

02

Calculating the mass of BaSO4

Substract the above result from the initial number of moles:

H2SO4 excess=35×10-3L H2SO4 soln×0.21 mol H2SO41 L H2SO4 soln-0.049 mol H2SO4                                  =2.45×10-2 mol.

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