Tums tablets contain CaCO3 that reacts with stomach acid as follows:

CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g)

 Suppose the tablet was analyzed by adding 50.00 mL of 0.200 M HCl, which resulted in leftover
excess HCl. The excess HCl was then analyzed by titrating with 20.00 mL of 0.100 M
NaOH. Calculate the mass of CaCO3 neutralized in this process.

Use the CaCO3 equation you have and add this one for the titration.

NaOH + HCl ==> NaCl + H2O

mols HCl initially = M x L = 0.01
excess HCl = mols NaOH = 0.002
mols HCl used for the tablet = 0.008
0.008 mols HCl x (1 mol CaCO3/2 mols HCl) = 0.004 mols CaCO3
g CaCO3 = mols CaCO3 x molar mass CaCO3 = ?

To calculate the mass of CaCO3 neutralized in this process, we need to find the number of moles of HCl that reacted with the CaCO3, and then use the balanced chemical equation to determine the number of moles of CaCO3 reacted.

Let's break down the steps to find the answer:

Step 1: Find the number of moles of HCl that reacted with NaOH.
The balanced chemical equation for the reaction between HCl and NaOH is:
HCl (aq) + NaOH (aq) -> NaCl (aq) + H2O (l)

From the balanced equation, we can see that the molar ratio between HCl and NaOH is 1:1. So, the number of moles of HCl is equal to the number of moles of NaOH used in the titration.

Given: Volume of NaOH used = 20.00 mL = 0.02000 L
Concentration of NaOH = 0.100 M

Using the equation: Moles = Concentration * Volume (in L)
Moles of NaOH = 0.100 M * 0.02000 L = 0.00200 moles

Since the molar ratio between HCl and NaOH is 1:1, the number of moles of HCl is also 0.00200 moles.

Step 2: Find the number of moles of CaCO3 that reacted with HCl.
From the balanced chemical equation, we can see that the molar ratio between CaCO3 and HCl is 1:2. This means that one mole of CaCO3 reacts with two moles of HCl.

Given: Concentration of HCl = 0.200 M
Volume of HCl used = 50.00 mL = 0.05000 L

Using the equation: Moles = Concentration * Volume (in L)
Moles of HCl = 0.200 M * 0.05000 L = 0.01000 moles

Since the molar ratio between CaCO3 and HCl is 1:2, the number of moles of CaCO3 is half the number of moles of HCl. Therefore, the number of moles of CaCO3 is 0.01000 / 2 = 0.00500 moles.

Step 3: Calculate the mass of CaCO3 neutralized.
The molar mass of CaCO3 can be obtained from the periodic table:
Molar mass of Ca = 40.08 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol

Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol) = 100.09 g/mol

Mass of CaCO3 = Moles of CaCO3 * Molar mass of CaCO3
Mass of CaCO3 = 0.00500 moles * 100.09 g/mol = 0.500 g

Therefore, the mass of CaCO3 neutralized in this process is 0.500 grams.