How many grams of sodium carbonate are present after the reaction is complete?

In a reaction between pure acetic acid and sodium bicarbonate, 32.5 grams of sodium bicarbonate was added to 9?

• In a reaction between pure acetic acid and sodium bicarbonate, 32.5 grams of sodium bicarbonate was added to 9.00 mL of acetic acid. Which reactant is the limiting reactant in this reaction? Name of limiting reactant (acetic acid or sodium bicarbonate): How many moles of the limiting reactant are present in this reaction? Moles of limiting reactant = mol How many moles of carbon dioxide do you expect to be produced? Moles of carbon dioxide = mol CO2 If the temperature in the laboratory is 21.1oC and the atmospheric pressure is 0.933 atm, what volume would be occupied by the carbon dioxide produced in this reaction (in cm3)? Volume of carbon dioxide = cm3 What would be the circumference of a spherical balloon containing this amount of carbon dioxide gas? Circumference of balloon = cm

• Answer:

  NaHCO3 + CH3COOH ====> CH3COONa + HOH + CO2 The Molarity of concentrated Acetic Acid is 17.5 mol/L Acetic Acid moles ============= C = 17.5 mol/L V = 9.00 mL = 9 mL * [1 L / 1000 mL] = 0.009 L n = C*V n = 17.5 * 0.009 n = 0.1575 moles Moles of Sodium Bicarbonate ==================== 1 mol of NaHCO3 Na = 23 H = 1 C = 12 O3 = 48 1 mole = 23 + 1 + 12 + 48 = 84 grams 1 mole = 84 grams. x mole = 32.5 grams x = 0.3869 mole A] Limiting reagent Acetic Acid. B] 0.1575 moles of NaHCO3 are present. C] It's all 1:1. You would get 0.1575 mol of Carbon Dioxide [Same as the limiting reagent. This does not always work out so well.] D] V * P = n*R*T R = 08205746 L atm/oC * mol n = 0.1575 T = 21 + 273 = 294oK P = 0.933 atm V = 0.1575 * 08205746 L*atm/oC * mol * 294oK /0.933 V = 4.073 L V = 4073 cm^3 E] V = 4/3 * pi * r^3 4073 = 1.33333 * 3.14 * r^3 r ^3 = 972 r = 9.91 cm Circumfrance = 2 * pi * r Circumfrance = 62.25 cm