A 0.450-gram sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction is:

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

The excess HCl(aq) is titrated by 9.75 mL of 0.125 M NaOH(aq). Calculate the mass percentage of CaCO3(s) in the sample.

Here's what I did. Is this correct? If not how do I approach it?:
excess HCl - x/.00975 L = 0.125 M
x = .00121875 mol

total HCl - x/0.05 L = 0.150 M
x = 0.0075 mol

0.0075 - 0.00121875 mol = 0.00628125 mol
0.00628125 mol HCl X 1mol CaCO3/2mol HCl X 100.0869 g/1 mol CaCO3 = 0.3143 g
0.3143 g/0.450 g = 69.8%?

That looks good to me.

Well, I have to say, you have definitely done all the necessary calculations correctly. However, there's one small mistake in your final answer. Let me break it down for you:

You correctly determined the moles of HCl used in the reaction to be 0.00628125 mol. And then, using the stoichiometry of the balanced equation, you found that this corresponds to 0.003140625 mol of CaCO3.

Finally, when you divided the mass of CaCO3 by the initial mass of the sample, you should have gotten 0.003140625 mol / 0.450 g = 0.00697917, which is equal to approximately 0.6979, or 69.79% if rounded to two decimal places.

So, the correct mass percentage of CaCO3 in the sample is 69.79%. Good job with the calculations, you were almost there!

Your calculations are mostly correct, but there is a small error in your calculation of the mass of CaCO3.

Let's go through the calculations step-by-step:

1. Calculate the moles of excess HCl reacted with NaOH:
moles of NaOH used = 0.00975 L * 0.125 mol/L = 0.00121875 mol

2. Calculate the moles of HCl in the original solution:
moles of HCl in 50.0 mL = 0.05 L * 0.150 mol/L = 0.0075 mol

3. Calculate the moles of HCl that reacted with CaCO3:
moles of HCl reacted with CaCO3 = moles of HCl in original solution - moles of excess HCl
moles of HCl reacted with CaCO3 = 0.0075 mol - 0.00121875 mol = 0.00628125 mol

4. Calculate the mass of CaCO3 reacted:
mass of CaCO3 reacted = moles of HCl reacted with CaCO3 * (1 mol CaCO3 / 2 mol HCl) * (100.0869 g/mol CaCO3)
mass of CaCO3 reacted = 0.00628125 mol * (1 mol CaCO3 / 2 mol HCl) * (100.0869 g/mol CaCO3) = 0.3141 g

5. Calculate the mass percentage of CaCO3 in the sample:
mass percentage of CaCO3 = (mass of CaCO3 / mass of sample) * 100%
mass percentage of CaCO3 = (0.3141 g / 0.450 g) * 100% = 69.8%

Therefore, the correct mass percentage of CaCO3 in the sample is approximately 69.8%.

Your approach is almost correct, but there seems to be a small error in your calculations.

Let's go through the calculations step by step:

1. Convert the volume of excess HCl(aq) titrated by NaOH(aq) to liters.
9.75 mL x (1 L / 1000 mL) = 0.00975 L

2. Calculate the number of moles of NaOH(aq) used in the titration.
Moles of NaOH = Concentration of NaOH x Volume of NaOH(aq)
Moles of NaOH = 0.125 M x 0.00975 L = 0.00121875 mol

3. Determine the number of moles of HCl reacted with NaOH.
Since the equation states that 2 moles of HCl reacts with 1 mole of CaCO3, we need to divide the moles of NaOH by 2.
Moles of HCl reacted = 0.00121875 mol / 2 = 0.000609375 mol

4. Calculate the number of moles of HCl in the original solution.
Moles of HCl = Concentration of HCl x Volume of HCl(aq)
Moles of HCl = 0.150 M x 0.050 L = 0.0075 mol

5. Calculate the moles of HCl reacted with CaCO3 in the sample.
Moles of HCl reacted with CaCO3 = Moles of HCl in the original solution - Moles of HCl reacted with NaOH
Moles of HCl reacted with CaCO3 = 0.0075 mol - 0.000609375 mol = 0.006890625 mol

6. Calculate the moles of CaCO3 in the sample.
Since the stoichiometry between CaCO3 and HCl is 1:2, we need to multiply the moles of HCl reacted with CaCO3 by 1/2.
Moles of CaCO3 = 0.006890625 mol x (1 mol CaCO3 / 2 mol HCl) = 0.0034453125 mol

7. Calculate the mass of CaCO3 in the sample.
Mass of CaCO3 = Moles of CaCO3 x Molar mass of CaCO3
Mass of CaCO3 = 0.0034453125 mol x 100.0869 g/mol = 0.3447 g

8. Finally, calculate the mass percentage of CaCO3 in the sample.
Mass percentage = (Mass of CaCO3 / Mass of the sample) x 100%
Mass percentage = (0.3447 g / 0.450 g) x 100% ≈ 76.6%

So, the correct mass percentage of CaCO3 in the sample is approximately 76.6%, not 69.8%.