I am doing a lab on Creating An Effective Airbag. We need to find out how much of the base is needed, which is NaHCO3,if we have 2.25 M of HCl, which is the acid. How much of the sample do we need? and how do we find out how much we need?

HCl + NaHCO3 ==> NaCl + H2O + CO2

You don't have enough information to answer the question. You have M HCl and that's all. You must have the volume of HCl if you are to calculate the grams NaHCO3. I might point out that this would not be a very good airbag as it would splatter HCl all over the inside of the vehicle AS WELL AS THE OCCUPANTS. 2.25M HCl could cause extensive damage especially to the eyes.

To determine how much NaHCO3 (base) is needed to react with 2.25 M of HCl (acid), you will need to use the balanced chemical equation for the reaction between NaHCO3 and HCl. The balanced equation is as follows:

NaHCO3 + HCl -> NaCl + H2O + CO2

From the balanced equation, you can see that the mole ratio between NaHCO3 and HCl is 1:1. This means that for every 1 mole of HCl, you will need 1 mole of NaHCO3 to react completely.

To calculate the amount of NaHCO3 needed, you first need to convert the given concentration of HCl (2.25 M) to moles. This can be done using the equation:

moles of HCl = concentration (M) x volume (L)

Since we do not have a specific volume mentioned in your question, I will assume you have a volume of 1 liter.

moles of HCl = 2.25 M x 1 L = 2.25 moles

Now, since the mole ratio between NaHCO3 and HCl is 1:1, you will need an equal number of moles of NaHCO3. Therefore, you will need 2.25 moles of NaHCO3.

To convert moles to grams, you need to know the molar mass of NaHCO3, which is:

Molar mass of NaHCO3 = (1 mol of Na x atomic mass of Na) + (1 mol of H x atomic mass of H) + (1 mol of C x atomic mass of C) + (3 mol of O x atomic mass of O)

= (1 x 22.99 g/mol) + (1 x 1.01 g/mol) + (1 x 12.01 g/mol) + (3 x 16.00 g/mol)

= 84.01 g/mol

Now you can determine the mass of NaHCO3 needed by multiplying the moles of NaHCO3 by its molar mass:

mass of NaHCO3 = moles of NaHCO3 x molar mass of NaHCO3

= 2.25 moles x 84.01 g/mol

= 189.03 grams

Therefore, you will need 189.03 grams of NaHCO3 for this reaction.