In a reaction between pure acetic acid and sodium bicarbonate, 30.6 grams of sod
ID: 845688 • Letter: I
Question
In a reaction between pure acetic acid and sodium bicarbonate, 30.6 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 21.2oC and the atmospheric pressure is 1.02 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
Explanation / Answer
CH3COOH + NaHCO3 = CH3COONa + H2O + 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.
So for 30.6 grams
x = 30.6/84 = 0.3643 mole
A]
So here the limiting reagent is Acetic Acid
B]
0.1575 moles of NaHCO3 are present.
C]
From the chemically balanced equation, 1 mole of acetic acid gives1 mole CO2. Thus, 0.1575 moles of Carbon Dioxide will be formed.
D]
V * P = n*R*T
R = 0.082 L atm/oK * mol
n = 0.1575
T = 21.2 + 273 = 294.2 oK
P = 1.02 atm
V = 0.1575 * 0.082 L*atm/oK * mol * 294.2 oK /1.02
V = 3.725 L
V = 3725 cm^3
E]
V = 4/3 * pi * r^3
3725 = 1.33333 * 3.14 * r^3
r ^3 = 889.73
r = 9.62 cm
Circumfrance = 2 * pi * r
Circumfrance = 2 * 3.14 * 9.62 = 60.41 cm
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