What mass of baking soda (NaHCO3) would be required to neutralize 10.00mL of 12.1M muriatic acid remaining in a empty water bottle?

muriatic acid is HCl.

HCl + NaHCO3 ==> NaCl + H2O + CO2

mols HCl = M x L = ?
mols NaHCO3 = mols HCl since 1 mol HCl reacts with 1 mol NaHCO3.
Then g NaHCO3 = mols NaHCO3 x molar mass NaHCO3.

By M x L do you mean Molarity (12.1) x Liters or moles x Liters?

Nevermind. I am an idiot.

To determine the mass of baking soda required to neutralize the muriatic acid, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between baking soda (NaHCO3) and muriatic acid (HCl).

The balanced chemical equation for the reaction is:
NaHCO3 + HCl -> NaCl + H2O + CO2

From the balanced equation, we can see that the molar ratio between NaHCO3 and HCl is 1:1. This means that one mole of NaHCO3 reacts with one mole of HCl.

Given that muriatic acid has a concentration of 12.1M and a volume of 10.00mL, we can first calculate the number of moles of HCl using the formula:

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

Since the volume is given in mL, we need to convert it to liters by dividing by 1000:
10.00mL / 1000 = 0.01000 L

Now, we can calculate the number of moles of HCl:
moles of HCl = 12.1M x 0.01000 L = 0.121 moles

Since the molar ratio between NaHCO3 and HCl is 1:1, the number of moles of NaHCO3 needed to neutralize the HCl is also 0.121 moles.

Now, to calculate the mass of NaHCO3, we need to use its molar mass. The molar mass of NaHCO3 can be calculated by adding up the atomic masses of each element:

Na: 22.99 g/mol
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (three oxygens)

Molar mass of NaHCO3 = 22.99 + 1.01 + 12.01 + (16.00 x 3) = 84.00 g/mol

Finally, we can calculate the mass of NaHCO3 needed using the formula:

mass = moles x molar mass
mass = 0.121 moles x 84.00 g/mol

Therefore, the mass of baking soda required to neutralize 10.00mL of 12.1M muriatic acid is approximately 10.1644 grams.