Question

I need someone to double check this for me if it's not too much trouble:

I have to make an airbag for my chem lab.
I'll be reacting HCl with sodium bicarbonate.
All stoichiometric ratios are 1:1 so I'll spare you the equation.

I have all other variables, I'll be solving for number of moles with the ideal gas law equation to find out how much reactant I need. to produce the gas I want.

Here's the math:
PV=nRT
n=PV/RT
n=(1atm)x(0.12L)/(0.0821 L x atm/mol x K) x (298.15K)
n= 4.9x10E-3

So that's the amount of moles of CO2 I need to fill the bag up entirely.

Answer 1

I'll post the rest shortly, but is this right so far?

Answer 2

In the following procedure it will be described how to make a small scale functioning air bag by reacting hydrochloric acid (HCl) with sodium bicarbonate (NaHCO3) to form carbon dioxide gas (CO2).

The following variables have been determined:
Pressure = 1 atm
Volume* = 120 mL = 0.12L
Gas constant (R): 0.0821 L·atm/mol·K
Temperature = 298.15K = 25.0C

*For the purposes of this procedural outline a Ziploc sandwich bag with a volume of 120mL has been selected.

Chemical/balanced equation:

HCl(aq) + NaHCO3(s) → NaCl(aq) + H2O(l) + CO2(g)

In order to determine the amount of reactants needed to fill the bag to its exact maximum capacity we must work backwards, utilizing what we know to determine what we do not.

Firstly, we utilize the ideal gas law equation PV=nRT, where n = number of moles. We will solve for n, and that will identify the number of moles needed to fill the volume of the bag:

PV=nRT
n=PV/RT
n=(1atm)·(0.12L)/(0.0821 L·atm/mol·K)·(298.15K)
n= 4.9x10E-3 moles of CO2

Answer 3

Using this information we can translate the moles of CO2 into the mass of each reactant needed via molar mass and stoichiometric ratio:

4.9x10E-3 molCO2 · 1mol HCl/1mol CO2 · 36.46g HCl/1mol HCl = 0.18g HCl*

4.9x10E-3 molCO2 · 1mol NaHCO3/1mol CO2 · 36.46g NaHCO3/1mol NaHCO3 = 0.41g NaOH3

*1mL ≈ 1g ∴ 0.18mL HCl