In a reaction between pure acetic acid and sodium bicarbonate, 10.0 grams of sod

Question

In a reaction between pure acetic acid and sodium bicarbonate, 10.0 grams of sodium bicarbonate was added to 21.0 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?

How many moles of carbon dioxide do you expect to be produced?

If the temperature in the laboratory is 24.1oC and the atmospheric pressure is 0.921 atm, what volume would be occupied by the carbon dioxide produced in this reaction (in cm3)?

What would be the circumference of a spherical balloon containing this amount of carbon dioxide gas in centimeters?

Explanation / Answer

get the number of moles of each reactant to determine the limiting reactant.

let's first deal with the NaHCO3:
10 g *(1 mol/84.01 g)= 0.119 mol NaHCO3

for the acetic acid, we need the density to get the number of moles. the density of pure acetic acid is 1.049g/ml
21.0 ml*(1.049g/ml)*(1mol/60.05g)= 0.3668442964 mol CH3COOH

since sodium bicarbonate has less moles (and the reactants have 1:1 ratio), it is the limiting reactant.

now we can compute the moles of CO2. but since we can see from the equation that sodium bicarbonate and CO2 has a 1:1 ratio, the number of moles of the sodium bicarb=number of moles of CO2=0.119 mol CO2

the next set of questions deal with ideal gases i assume. we can therefore use the ideal gas equation PV=nRT. we already have the number of moles, n, the pressure, P=0.921 atm, the temperature, T, can be readily obtained (T(K)=24.1 + 273.15=297.25 K), and the most ideal gas constant to be used is 0.0821 because it has the units L-atm/mol-K.

V=nRT/P=3.369579824 L= 3.15*10^(3) cm^3

to get the circumference, we need the equation of volume of a sphere:
V=(4/3)*pi*r^3
get r=9.09 cm

the circumference is calculated as:
C=2*pi*r=57.10 cm

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