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

Question

After the reaction occurred, it took 15.3 mL of 0.7034 M NaOH to neutralize the excess acid.

a) How much HCl (in mL) of was neutralized by NaOH?

b) How much HCl (in mL) was needed to neutralize 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.

Answer 1

The problem is a little more complicated because it wants mL (not mols or millimols).
a. Excess acid = millimols NaOH = mL x M = 15.3 x 0.7034 M NaOH. I'll leave it to you to calculate how many mL HCl that is.

b. mL HCl used in the neutralization was 30 mL - mL from part a = ?

c. CaCO3 x 2HCl ==> CaCl2 + H2O + CO2
You used ? mL HCl (from part b) and that x M = millimols HCl used in the neutralization. Convert to mols. Then mols CaCO3 = mols HCl/2 from the stoichiometry. grams CaCO3 = mols CaCO3 x molar mass CaCO3.

d. %CaCO3 = (g CaCO3/mass sample)*100 = ?

Answer 2

so here are the answers for
a)
NaOH + HCl → NaCl + H2O
(15.3 mL) x (0.7034 M NaOH) x (1 mol HCl / 1 mol NaOH) /
(0.831 M HCl) = 12.95 mL HCl

b)
CaCO3 + 2 HCl → CaCl2 + CO2 + H2O
(30.0 mL HCl total) - (12.95 mL HCl for NaOH) = 17.05 mL HCl for the CaCO3

how would I solve the others?

Answer 3

c.
mols HCl = M x L = ?
mols CaCO3 = 1/2 that
grams CaCO3 = mols x molar mass = ?

d. See my other response and plug in the numbers.