Question

CH3COOH + NaHCO3 --> H2O + CHCOONa + CO2

Question 1.) The moles of acetic acid in 70.0 mL of 0.833 M acetic acid solution is ___ moles. <-- 0.0583 moles ?

The formula mass of NaHCO3 is 84.01 g/mol. Use this to calculate the moles of NaHCO₃ below:
Trial 1:  2.40 g NaHCO₃ contains ___ moles? <-- 0.0286 moles ?
Trial 2: 3.60 g NaHCO₃ contains___ moles? <-- 0.0429 moles ?
Trial 3: 4.80 g NaHCO₃ contains ___ moles? <-- 0.0571 moles ?
Trial 4: 6.00 g NaHCO₃ contains ___ moles? <-- 0.0714 moles ?
Trial 5: 7.20 g NaHCO₃ contains ___ moles? <-- 0.0857 moles ?

Part 1.A.) Determine which reactant limits the amount of carbon dioxide produced in each bottle from the previous question. Type ONLY the following options into each blank:  NaHCO3 or  CH3COOH or neither. Subscripts for the formulas are NOT needed.

Trial 1: ____

Trial 2: ____

Trial 3: ____

Trial 4: ____

Trial 5: ____

Part 1.B.) Select the image that best represents how the balloons will look after the reaction is complete. Pay close attention to the sizes of the balloons.

Choose A.),  B.), C.), D.)

Part 1.C.) Explain the reasoning behind the picture you chose for the previous question.

Answer 1

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Question 1.)

The moles of acetic acid in 70.0 mL of 0.833 M acetic acid solution is:

First change mL to L

And use the equation , rewrite the equation as:

And substitute values:

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Calculating the moles of NaHCO₃:

On each trial we know the mass of NaHCO₃, so we can use the molecular mass to calculate the number of moles:

Trial 1:             

Trial 2:             

Trial 3:             

Trial 4:             

Trial 5:             

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Part 1.A.)

First balance the chemical equation as:

3CH₃COOH + 4NaHCO₃ -> 6H₂O + 4CHCOONa + 4CO₂

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We know that there are 0.0583moles of acetic acid

So we are going to be adding NaHCO₃ moles, we can represent it in a table:

	3CH3COOH	+ 4NaHCO3	->	6H2O + 4CHCOONa + 4CO2
Before the adition	0.0583moles
Adition		?
After the adition	?	?

Because the stoichiometric ratio of the reaction is 3:4, we need to multiply by 3/4 the moles of NaHCO₃

After multiply by ¾, we can compare the values of moles of NaHCO₃ with the number of moles of acetic acid.

+The reagent that will limit the amount of carbon dioxide produced, is the reagent with the least amount of moles:

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Trial 1: 0.02856 moles of NaHCO₃ :

Reactant	CH3COOH	NaHCO3
Moles	0.0583	0.02142

The limiting reactant will be NaHCO₃ because it has the least amount of moles.

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Trial 2: 0.04285 moles of NaHCO₃ :

Reactant	CH3COOH	NaHCO3
Moles	0.0583	0.0321

The limiting reactant will be NaHCO₃ because it has the least amount of moles.

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Trial 3: 0.05713 moles of NaHCO₃ :

Reactant	CH3COOH	NaHCO3
Moles	0.0583	0.0428

The limiting reactant will be NaHCO₃ because it has the least amount of moles.

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Trial 4: 0.07142 moles of NaHCO₃ :

Reactant	CH3COOH	NaHCO3
Moles	0.0583	0.0535

The limiting reactant will be NaHCO₃ because it has the least amount of moles.

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Trial 5: 0.085704 moles of NaHCO₃ :

Reactant	CH3COOH	NaHCO3
Moles	0.0583	0.0642

The limiting reactant will be CH₃COOH because it has the least amount of moles

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