READ THE EXPERIMENTAL DISCUSSION FIRST REPORT ALL ANSWERS TO THE CORRECT SIGNIFI
ID: 1030821 • Letter: R
Question
READ THE EXPERIMENTAL DISCUSSION FIRST REPORT ALL ANSWERS TO THE CORRECT SIGNIFICANT FIGURES SHOW CALCULATION SET-UP on the next page. PART B: Molarity Determination via Solution Stoichiometry 7.625 g 7.976 g o-351 s Mass of Empty 50 mL Beaker Mass of Beaker and NaCl (final heating) Mass of NaCI Produced (Experimental Yield of NaCI) Moles of Na,CO, Consumed Volume of Na,CO, Solution Molarity of Na,CO, Solution PART C: Reaction Stoichiometry Mass of Empty 50 mL. Beaker Mass of Beaker and Unknown Solid Mass of Unknown Solid Mass of Beaker and NaCl (final heating) Mass of NaCI Produced (Experimental Yield of NaCI) Theoretical Yield of NaCl if Unknown Solid is Na,CO, Theoretical Yield of NaCl if Unknown Solid is NaHCO Formula of the Unknown Solid Percent Yield moles 0.00500 Liter 8.125 g 8.387 g 0-262 8.405 g grams NaCI grams NaCl 2Explanation / Answer
Solution:
PART B) NaCl can be produced from sodium carbonate by the reaction between sodium carbonate and hydrochloric acid. Here the formation of weaker acid is favoured resulting in the production of the neutral salt and carbonic acid. The balanced equation for this reaction is given as: Na2CO3 + 2HCl -----> 2NaCl + H2CO3. From the balanced equation it can be seen that for every mole of carbonate consumed, two moles of sodium chloride is formed. Thus from the mass of sodium chloride, by calculating the moles of the salt formed, we can conclude that half that many moles of sodium carbonate was consumed.
Now, from the mass difference of empty beaker and that of the beaker with the salt, the mass of NaCl is found as 7.976g - 7.625g = 0.351g. The molar mass of NaCl is 58.4g/mol and so the no.of moles in 0.351g is given as 0.351/58.4 = 6.0102mmol (1mmol = 0.001mol) implying that the no.of moles of sodium carbonate consumed is 6.0102/2 = 3.0051mmol.
Molarity is defined the the no.of moles of a substance dissolved in 1L of solution. Thus the molarity of sodium carbonate in the 0.005L of solution used can now be calculated from the no.of moles of the carbonate salt as molarity = moles of sodium carbonate*(1/0.005) = 0.60102M solution of sodium carbonate.
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