4. A 5.309 g antacid tablet, with CaCO3 as the active ingredient, was mixed was 30.0 mL of 0.831 M HCl.

It took 15.3 mL of 0.7034 M NaOH to neutralize the excess acid.
a) Calculate the volume of HCl neutralized by the NaOH.
b) Calculate the amount of HCl (in mL) neutralized by the antacid.
c) Calculate the mass of CaCO3 (in g) that reacted with HCl.
d) Calculate the mass percent of CaCO3 present in the antacid

To solve this problem, we will use the concept of stoichiometry and the principles of balanced chemical equations. Here are the steps to find the answers to each part:

a) To calculate the volume of HCl neutralized by NaOH, we need to determine the number of moles of HCl that reacted with NaOH. First, we calculate the number of moles of NaOH:

Moles of NaOH = concentration of NaOH * volume of NaOH
= 0.7034 M * 0.0153 L (convert mL to L)
= 0.0108 mol NaOH

Since the reaction between HCl and NaOH is a 1:1 ratio, the number of moles of HCl neutralized is also 0.0108 mol.

Now, we can calculate the volume of HCl neutralized by dividing the moles of HCl by its concentration:

Volume of HCl neutralized = moles of HCl / concentration of HCl
= 0.0108 mol / 0.831 M (convert mL to L)
= 0.012998 L (convert back to mL)
= 12.998 mL (rounded to 13 mL)

Therefore, the volume of HCl neutralized by NaOH is 13 mL.

b) To calculate the amount of HCl neutralized by the antacid, we need to find the difference between the initial volume of HCl and the volume of HCl neutralized by NaOH. The initial volume of HCl is given as 30.0 mL, so:

Amount of HCl neutralized by the antacid = Initial volume of HCl - Volume of HCl neutralized by NaOH
= 30.0 mL - 13 mL
= 17 mL

Therefore, the amount of HCl neutralized by the antacid is 17 mL.

c) In this step, we will calculate the moles of CaCO3 that reacted with the HCl. Since the reaction between CaCO3 and HCl is also a 1:1 ratio, the number of moles of CaCO3 is equal to the number of moles of HCl that were neutralized:

Moles of CaCO3 = Moles of HCl neutralized by the antacid
= 0.0108 mol

Now, we can calculate the mass of CaCO3 using its molar mass:

Mass of CaCO3 = Moles of CaCO3 * Molar mass of CaCO3
= 0.0108 mol * 100.0869 g/mol (obtained from the periodic table)
= 1.081 g

Therefore, the mass of CaCO3 that reacted with HCl is 1.081 g.

d) Finally, to calculate the mass percent of CaCO3 present in the antacid, we need to determine the mass of CaCO3 in the entire antacid tablet. The mass of the tablet is given as 5.309 g, so:

Mass percent of CaCO3 = (Mass of CaCO3 / Mass of antacid tablet) * 100%
= (1.081 g / 5.309 g) * 100%
= 20.36 %

Therefore, the mass percent of CaCO3 present in the antacid tablet is 20.36%.