Stoichiometry Lab Reactions: 5 All 5 reactions performed in a 500 mL bottle Mass

Question

Stoichiometry Lab

Reactions: 5

All 5 reactions performed in a 500 mL bottle

Mass of baking soda added: 1, 3, 5, 7, and 9.00 grams

100 mL of vinger added all 5 reactions (5%)

ablesBoutque Dental Hyalone-L.ccc M McGraw-Hill Conne... Dmynw.aduLaramie County Co Volume Moles of Mass of acetic baking soda Moles of sodium Diameter vinegar acid hydrogen of the Visual Observations added added added Carbonate added balloon 2 3 4 5 In the Calculations section, show calculations to find each of the following: Mass of acetic acid added to the first bottle knowing that the % acidity for vinegar is a mass percent. That is, 5% acidity vinegar is 5 g acetic acid/100 g vinegar, and 4% acidity vinegar is 4 g acetic acid/100 g vinegar. · .Molar mass of acetic acid, C2H402. Moles of acetic acid added to the first bottle. Molar mass of sodium hydrogen carbonate, NaHCO3. Moles of sodium hydrogen carbonate added to the first bottle. (You will need to calculate moles of acetic acid and sodium hydrogen carbonate for the other bottles as well, but you do not need to show those calculations here.) In your Conclusions, give the balanced equation for the reaction of acetic acid with sodium hydrogen carbonate. Be sure to use appropriate superscripts and subscripts and include the phase symbol (s, t g, aq) for each reactant and product in the balanced equation. Note the reaction (bottle #) that had the correct ratio to match the balanced equation. In which bottles was acetic acid limiting reactant? In which bottles was sodium hydrogen carbonate limiting reactant? Comment on how well your balloon diameter measurements reflect these limiting reactants

Explanation / Answer

Reaction of acetic acid with baking soda

CH3COOH(aq) + NaHCO3(s) ---> CH3COONa(aq) + H2O(l) + CO2(g)

test reaction 1,

mass NaHCO3 = 1 g

moles NaHCO3 = 1 g/84 g/mol = 0.012 moles

moles acetic acid needed = 0.012 moles

vinegar solution = 5%

molar mass acetic acid = 60.05 g/mol

5 g acetic acid in 100 ml water

molarity of vinegar solution = 5 g/60.05 g/mol x 0.1 L = 0.833 M

actual volume of vinegar solution needed = 0.012 moles x 1000/0.833 M = 14.40 ml

vinegar = excess reagent

limiting reactant : NaHCO3

test reaction 2,

mass NaHCO3 = 3 g

moles NaHCO3 = 3 g/84 g/mol = 0.036 moles

moles acetic acid needed = 0.036 moles

actual volume of vinegar solution needed = 0.036 moles x 1000/0.833 M = 43.22 ml

vinegar = excess reagent

limiting reactant : NaHCO3

test reaction 3,

mass NaHCO3 = 5 g

moles NaHCO3 = 5 g/84 g/mol = 0.059 moles

moles acetic acid needed = 0.059 moles

actual volume of vinegar solution needed = 0.059 moles x 1000/0.833 M = 70.83 ml

vinegar = excess reagent

limiting reactant : NaHCO3

test reaction 4,

mass NaHCO3 = 7 g

moles NaHCO3 = 7 g/84 g/mol = 0.083 moles

moles acetic acid needed = 0.083 moles

actual volume of vinegar solution needed = 0.083 moles x 1000/0.833 M = 100 ml

vinegar : NaHCO3 = 1 : 1

test reaction 5,

mass NaHCO3 = 9 g

moles NaHCO3 = 9 g/84 g/mol = 0.107 moles

moles acetic acid needed = 0.107 moles

actual volume of vinegar solution needed = 0.107 moles x 1000/0.833 M = 128.45 ml

NaHCO3 = excess reagent

limiting reactant : Vinegar

Observation : In each of the reaction, evolution of CO2 gas (bubbles) will be seen.Acid-base reaction is exothermic reaction. The solution becomes hot after the reaction.

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