Chemistry help on airbag lab for finding moles theoretical yield

Question

If you made an air bag using 1.000 g baking soda and 1.500 mL 3.00 M acetic acid what would the theretical yield of CO2 gas in moles? which reactant would be limiting?

NaHCO3+CH3COOH=CO2+CH3COONa+H2O

and what would the volume of CO2 in mL be if the temp of the room is 20.3 C and the pressure 1.01 atm?

Answer 1

Baking soda = NaHCO₃ (sodium bicarbonate)

Acetic acid = CH₃COOH

NaHCO₃ + CH₃COOH ==> CO₂ + CH₃COONa + H₂O ... balanced equation

An easy way to find the limiting reactant is to divide the moles of each reactant by the corresponding coefficient in the balanced equation.

For NaHCO3: 1.000 g NaHCO3 x 1 mol NaHCO3 / 84.00 g = 0.01190 mols (÷1->0.01190)

For CH3COOH: 1.500 ml x 1 L / 1000 ml x 3.00 mol/L = 0.004500 mols (÷1->0.0045)

Since 0.0045 is less than 0.0119, CH3COOH IS LIMITING

Use the mols of the limiting reactant to find the mols of CO2 (theoretical yield in mols):

0.004500 mols CH3COOH x 1 mol CO2 / mol CH3COOH = 0.004500 mols CO2

To find the volume of CO2 (in mls) @20.3ºC and 1.01 atm pressure, use the ideal gas law:

PV = nRT

P = pressure = 1.01 atm

V = volume in L = ?

n = mols = 0.004500 mols

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 20.3ºC + 273.15 = 293.5K

Solving for V ...

V = nRT/P = (0.0045)(0.0821)(293.5) / 1.01

V = 0.107 L x 1000 ml / L = 107 ml