Calculations Sheet wart 1: Citric Acid + Sodium Hydrogen Carbonate Parstrategy w

Question

Calculations Sheet wart 1: Citric Acid + Sodium Hydrogen Carbonate Parstrategy with this set of calculations is to determine the amount of heat transfered per one mole of the limiting reagent in the experiment. The limiting reagent in Part 1 is sodium hydrogen carbonate. To determine the amount of heat transferred in the reaction we use AH- -mSAT. Oace this value has been calculated then the units are converted from Joules to kilojoules. Thermochemistry Volume of citric acid solution Density of citric acid solution Mass of citric acid (d-m/V) Specific heat capacity of citric acid solution Change in temperature for reaction ?? (-msAT) Mass NaHCO, (g) Moles NaHCO, (mol)* AH per mole of NaHCO3 [??- moles NaHCO ] AH per 3 moles of NaHCO, *. d. mol J-mol kJ mol AH per 3 moles of NaHCO, in kJ/mol (1 kJ- 1000 J) Use the mass and molar mass of NaHCO **There are three moles of Naldo, in the balanced equation, so ?? per mol Nalco, is multiplied by 3. H,C,H, O,(aq) +3NaHCO,(s) 3C0,(8)+3H,00)+ Na,C,H,O, (ag)

Explanation / Answer

Part 1: citric acid + 3NaHCO3

dH = -(4.0 ml x 1.02 g/ml) x 4.68 J/g.oC x 15.4 oC

      = -294.054 J

moles NaHCO3 reacted = 3 x 0.0408 g/192.124 g/mol = 0.00064 mol

mass NaHCO3 = 0.00064 mol x 84 g/mol = 0.054 g

dH per mole of NaHCO3 = -294.054 J/0.00064 mol x 1000 = -459.46 kJ/mol

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Part 2: Mg + 2HCl

dH = -(4.0 ml x 1.02 g/ml) x 4.68 J/g.oC x 8 oC

      = -152.76 J

moles Mg reacted = 0.0408 g/2 x 36.5 g/mol = 0.00056 mol

mass Mg = 0.00056 mol x 24.305 g/mol = 0.014 g

dH per mole of HCl = -152.76 J/0.00056 mol x 1000 = -272.8 kJ/mol

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