How many grams of baking soda, NaHCO3, are needed to react with 125 mL of stomach acid having an HCl concentration of 0.05984 M?

The reaction is

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

Do I need to chenage mL to L first? I'm really stuck with this question. I don't know where to start.

Thanks for letting us know.

Yes, it is a good idea to convert the volume from milliliters (mL) to liters (L) before starting the calculation. This will ensure that the units are consistent throughout the calculation.

To convert 125 mL to L, you divide the volume in mL by 1000:

125 mL ÷ 1000 = 0.125 L

Now that you have the volume in liters, you can proceed with the calculation.

Yes, you will need to convert milliliters (mL) to liters (L) in order to correctly calculate the amount of baking soda required.

To begin solving the problem, you need to first determine the number of moles of HCl in the solution. Moles can be calculated using the formula:

moles = concentration (mol/L) x volume (L)

Given:
- Concentration of HCl = 0.05984 M
- Volume of stomach acid = 125 mL

Step 1: Convert mL to L
125 mL x (1 L / 1000 mL) = 0.125 L

Step 2: Calculate moles of HCl
moles of HCl = 0.05984 M x 0.125 L = 0.00748 moles

The balanced equation tells us that the mole ratio of NaHCO3 to HCl is 1:1. Therefore, the number of moles of NaHCO3 required will also be 0.00748 moles.

Finally, to determine the mass of NaHCO3 required, you need to use the molar mass of NaHCO3, which is 84.01 g/mol.

Step 3: Calculate mass of NaHCO3
mass of NaHCO3 = moles of NaHCO3 x molar mass of NaHCO3
mass of NaHCO3 = 0.00748 moles x 84.01 g/mol = 0.628 g

Therefore, you will need approximately 0.628 grams of baking soda (NaHCO3) to react with 125 mL of stomach acid with an HCl concentration of 0.05984 M.

I can solve it now. Sorry about the post.