An analysis of generic antacid tablets labeled to contain 250mg of CaCO3 (MM= 100.09g/mol) per tablet of active ingredient was performed. The analysis was performed by dissolving a 0.550g sample of the antacid tablet in 75.00mL of 0.1055M HCl. The excess acid was back-titrated with exactly 44.97mL of 0.125M NaOH. The average weight of a tablet is 1.045g. The tablet came from a bottle of 125 tablets that cost $3.49.

Calculate the actual mg per tablet of CaCO3 in the generic antacid.

I will estimate all of the numbers which means you must recalculate all of them for better accuracy.

millimols HCl added initially = mL x M = about 7.9 millimols HCl.
millimols NaOH to backtitrate = mL x M = about 5.6 millimoles.
millimoles HCl used by the tablet when disolved = 7.9-5.6 = about 2.3.
CaCO3 + 2HCl ==> CaCl2 + CO2 + H2O
Convert 2.3 millimoles HCl to millimoles CaCO3 so 2.3/2 = 1.2 millimoles CaCO3 in the sample or about 0.0012 mols.
g CaCO3 = mols CaCO3 x molar mass CaCO3 and that x 1000 = about 120mg CaCO3 in the 0.550 g sample. Convert to mg in a single table with the average weight of 1.045 by about 120 mg x (1.045/0.550) = ?
Post your work if you get stuck.

To calculate the actual mg per tablet of CaCO3 in the generic antacid, we need to follow these steps:

Step 1: Determine the amount of HCl that reacts with the CaCO3.
We can use the balanced chemical equation between HCl and CaCO3 to find the moles of HCl that react with our sample mass of CaCO3:

2 HCl + CaCO3 -> CaCl2 + H2O + CO2

From the balanced equation, we can see that 2 moles of HCl react with 1 mole of CaCO3. So, we need to convert the mass of CaCO3 to moles by dividing the mass by the molar mass:

Mass of CaCO3 = 250 mg = 0.250 g
Molar mass of CaCO3 = 100.09 g/mol

Moles of CaCO3 = Mass of CaCO3 / Molar mass of CaCO3
= 0.250 g / 100.09 g/mol

Step 2: Determine the moles of HCl used.
We have the molarity and volume of the HCl solution used for the analysis. Using the equation Molarity (M) = Moles / Volume (L), we can rearrange it to find the moles:

Moles of HCl = Molarity of HCl x Volume of HCl
= 0.1055 mol/L x 0.07500 L

Step 3: Determine the moles of NaOH used.
We have the molarity and volume of the NaOH solution used for back-titration. Using the same equation, we can calculate the moles of NaOH:

Moles of NaOH = Molarity of NaOH x Volume of NaOH
= 0.125 mol/L x 0.04497 L

Step 4: Calculate the excess moles of HCl.
To find the excess moles of HCl, we need to subtract the moles of NaOH used from the moles of HCl used:

Excess moles of HCl = Moles of HCl - Moles of NaOH

Step 5: Find the moles of CaCO3 reacted.
Since 2 moles of HCl react with 1 mole of CaCO3, we can relate the moles of CaCO3 to the moles of HCl reacted:

Moles of CaCO3 reacted = (Excess moles of HCl) / 2

Step 6: Calculate the mg per tablet of CaCO3.
Finally, we can use the moles of CaCO3 reacted to find the actual mg of CaCO3 per tablet:

Mg per tablet of CaCO3 = (Moles of CaCO3 reacted x Molar mass of CaCO3) / Number of tablets

Number of tablets = Total mass of the bottle / Average weight of a tablet
= 125 tablets

Now, substitute the values into the equation:

Mg per tablet of CaCO3 = (Moles of CaCO3 reacted x Molar mass of CaCO3) / Number of tablets
= (Moles of CaCO3 reacted x 100.09 g/mol) / 125 tablets

Please substitute the values into the equation to calculate the answer.