Instructions: Complete the following questions and hand in at the start of your

Question

Instructions: Complete the following questions and hand in at the start of your lab period. Show your work with units and correct significant figures for all questions that involve a calculation Consult the amounts used in part III of the procedure. Calculate how many millimoles of KI is in excess to prove that KIO, is the limiting reactant in equation 4. No work, no credit. 1. 2. How many grams of sodium thiosulfate pentahydrate, Na,S,0,.5H,0, are needed to make 500.0 mL of 0.080M- Na2S,03? Show your work! molar mass Nazmo3:159,775 ginal

Explanation / Answer

The balanced chemical equation for the reaction:

1) KIO3(aq)+6H+(aq)+5I-(aq)----->3I2(aq)+3H2O(l)+K+(aq)

2) C6H8O6(aq)+I2(aq)--->C6H6O6(aq)+2I-(aq)+2H+(aq)

C6H8O6,ascorbic acid or vit C

Q3)mol KIO3=0.040L*0.02912mol/L=0.00116 mol

mol KI=2g/166.003g/mol=0.012 mol

KI and KIO3 react in molar ratio of 5:1 according to the balanced quation,

So mol(KI)/mol(KIO3)=5:1

mol(KI)=5*mol(KIO3)=5*0.00116 mol=0.0058

excess KI=0.012mol-0.0058mol=0.0062 mol=0.0062mol*(10^-3mmol/mol)=6.2mmol

excess KI=6.2mmol

Thus,KI is the excess reactant and KIO3 is the limiting reactant

2) molar mass of Na2S2O3.5H2O=248.184 g/mol

mol of Na2S2O3.5H2O=0.5L*0.08 mol/L=0.04mol

mass of Na2S2O3.5H2O=mol*molar mass=0.04mol*(248.184 g/mol)=9.927 g

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