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Chemical Reaction of Sodium Acetate and Water

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Carola Lama

on 21 November 2013

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Transcript of Chemical Reaction of Sodium Acetate and Water

Sodium Acetate Lab
Chemical Reaction of Sodium Acetate and Water
NaC2H3O2 + H2O Na(OH) + HC2H3O2
Atomic Representation of the Reaction
NaC2H3O2 + H2O Na(OH) + HC2H3O2

Lab Representation of the Reaction
Identifying the Limiting Reactant
NaC2H3O2 23+12(2)+1(3)+16(2)= 82 g/mol

20 g/82 g/mol=
0.24 mol NaC2H3O2
During the double replacement reaction in which sodium acetate and water take place, the sodium separates from the acetate and bonds with hydroxide. Also, H20 is treated as HOH, which makes one hydrogen bond with acetate.
The mass of the reactants (NaC2H3O2 + H2O) decrease, while the products' (Na(OH) + HC2H3O2) mass increase.
H2O 1(2)+16= 18 g/mol

20 ml/18 g/mol=
1.11 g/mol H20
Since the reactants don't have coefficients, it is unnecessary to divide the moles by anything, making
sodium acetate the limiting reactant
water the excess
Step 1: Add 20 mm of water and 20 grams of sodium acetate into a flask.
Mass of Na(OH)
Mass= molar mass X moles
Moles of limiting reactant: 0.24
Na(OH) 23+16+1= 40 g/mol
40 g/mol Na(OH) X 0.24 mol NaC2H3O2=

9.6 g Na(OH)
Step 2: Use the bunsen burner to heat the reactants until the solution is liquid.
Step 3: Let it cool down in cool water outside of the solution.
Step 4: Take the flask out and dry it. Afterwards, pour the liquid into the cristallized sodium acetate, slowly.
Step 5: You will get the product, supersaturated, which will be hot and turn into solid.
Percentage error
FORMULA: actual-theoretical/theoretical X 100
Actual mass of product: 29.5 grams
29.5-19/19 X 100=
What might have occurred during the lab that made the numbers higher than they should was that the measurements were not exact, or some of the solution might have spilled during the reaction.
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