A 1.605g sample of dry calcium carbonate and calcium chloride mixture was dissolved in 35.00mL of 1.000mol-dm-3 hydrochloric acid solution. The only portion of the mixture that reacts with the hydrochloric acid solution is the calcium carbonate. What was the calcium chloride percentage in the original sample, if 31.51mL of 0.0950mol-dm-3 sodium hydroxide was used to titrate excess hydrochloric acid?

I don't know where to go from here:
CaCO3 + 2HCl -> CaCl2 + CO2 +H2O
NaOH +HCl -> H2O + NaCl

Please help!! :)

mmol = millimole

mmol HCl initially = mL x M = 35.00 x 1.00 = 35.00.
mmols NaOH used to titrate excess is 31.51 x 0.0950 = 3 approx but you need to do a more accurate calculation. All of the other numbers that follow are approximations, too, so you need to redo all of this more accurately.

mols HCl used in the reaction with CaCO3 = Initial - final = approx 35 - 3 = 32 or 0.032 mols HCl
Convert that to mols CaCO3 and that's 0.032 x (1 mol CaCO3/2 mols HCl) = 0.016 mols CaCO3.
Than g CaCO3 = mols CaCO3 x molar mass = g CaCO3
1.605 g mixture - grams CaCO3 = grams CaCl2.
%CaCl2 = (g CaCl2/1.605)*100 = ?
Post your work if you get stuck. I think the percent CaCl2 is a small number.
g CaCO3 = 0.032 x molar mass = approx 3.2

Thanks, this helped me figure out what I needed to do, I got 0.25% CaCl2

Ignore the last line. I don't know how it got posted; it shouldn't be there. The rest of the problem is ok.

To find the percentage of calcium chloride in the original sample, we need to calculate the amount of calcium carbonate that reacted with hydrochloric acid and the amount of calcium chloride formed.

Let's break down the problem step by step:

Step 1: Calculate the number of moles of hydrochloric acid used.
We're given the volume of the hydrochloric acid solution is 35.00 mL, and its concentration is 1.000 mol/dm³ (or 1.000 mol/L). We can use the formula:

moles of HCl = concentration × volume (in dm³)

moles of HCl = 1.000 mol/dm³ × (35.00 mL / 1000 mL/dm³) = 0.035 mol

Step 2: Calculate the number of moles of sodium hydroxide used in the titration.
We're given the volume of sodium hydroxide (NaOH) solution is 31.51 mL and its concentration is 0.0950 mol/dm³. Using the same formula as above:

moles of NaOH = concentration × volume (in dm³)

moles of NaOH = 0.0950 mol/dm³ × (31.51 mL / 1000 mL/dm³) = 0.002996 mol

Step 3: Calculate the number of moles of HCl reacted with NaOH.
From the balanced equation NaOH + HCl → H2O + NaCl, we can see that 1 mole of NaOH reacts with 1 mole of HCl. Therefore, the moles of HCl reacted with NaOH is also 0.002996 mol.

Step 4: Determine the excess moles of HCl.
Since only the calcium carbonate reacts with HCl, the moles of calcium carbonate and moles of HCl reacted should be in a 1:2 ratio:

moles of CaCO3 reacted = 2 × moles of HCl reacted = 2 × 0.002996 mol = 0.005992 mol

Step 5: Calculate the moles of CaCl2 formed.
From the balanced equation CaCO3 + 2HCl → CaCl2 + CO2 + H2O, we can see that 1 mole of CaCO3 reacts to form 1 mole of CaCl2. Therefore, the moles of CaCl2 formed is also 0.005992 mol.

Step 6: Calculate the mass of CaCl2 formed.
The molar mass of CaCl2 is approximately 110.98 g/mol. Thus, mass = moles × molar mass:

mass of CaCl2 = 0.005992 mol × 110.98 g/mol ≈ 0.665 g

Step 7: Determine the percentage of CaCl2 in the original sample.
We're given the mass of the dry mixture was 1.605 g. Therefore, the percentage of CaCl2 = (mass of CaCl2 / mass of sample) × 100%:

percentage of CaCl2 = (0.665 g / 1.605 g) × 100% ≈ 41.38%

Therefore, the calcium chloride percentage in the original sample is approximately 41.38%.