In a reaction between pure acetic acid and sodium bicarbonate, 15.5 grams of sod

Question

In a reaction between pure acetic acid and sodium bicarbonate, 15.5 grams of sodium bicarbonate was added to 24.0 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 22.1 degree C and the atmospheric pressure is 0.954 atm, what volume would be occupied by the carbon dioxide produced in this reaction (in cm3)? Volume of carbon dioxide = cm^3 What would be the circumference of a spherical balloon containing this amount of carbon dioxide gas? Circumference of balloon = cm

Explanation / Answer

CH3COOH + NaHCO3 --> NaCH3COO + CO2 + H2O

Now according to the equation

1 mol of CH3COOH reacts with 1 mol of NaHCO3 but we have to calculate the moles as given

moles of NaHCO3 = given mass / molar mass = 15.5 / 84 = 0.184 moles

Now ,

density of pure acetic acid = 1.049 g/mL

So, mass = density * volume = 1.049 * 24 = 25.2 g

So, moles = 25.2 / 60 = 0.42 moles

Now,

1 mol of CH3COOH reacts with 1 mol of NaHCO3 but as per calculation we have 0.184 moles of NaHCO3 reacting with 0.42 moles of CH3COOH .

So , NaHCO3 is in limited amount hence it is limiting reagent.

moles of limiting reagent = 0.184

1 mole of NaHCO3 produces 1 mol of CO2

So, 0.184 mol of NaHCO3 produces 0.184 mol of CO2 .

Using Ideal gas Law

PV = nRT

T = 273.15 + 22.1 = 295.25 K

R = 0.0821 L-atm/mol-K

0.954*V = 0.184*0.0821*295.25

V = 4.68 L = 4.68 x 10^3 cm^3

Volume of balloon = 4/3*pi*r^3

4.68 x 10^3 = 4/3 * 3.14 * r^3

So ,

r comes out to be 10.38 cm

Hence

Circumference of spherical balloon = 2*pi*r = 2*3.14*10.38 = 65.2 cm

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