A 5.309 g antacid tablet, with CaCO3 as the active ingredient, was mixed was 30.0 mL of 0.831 M HCl.
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.
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 = ?
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?
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.
To solve these problems, we need to use the concept of stoichiometry and the principles of acid-base neutralization. Let's go step by step to find the answers to each question:
a) To calculate the amount of HCl neutralized by NaOH, we need to determine the excess amount of HCl that remains after the reaction. We know that 0.7034 M NaOH was used to neutralize the excess acid.
First, calculate the moles of NaOH used:
moles of NaOH = concentration of NaOH (M) x volume of NaOH (L)
moles of NaOH = 0.7034 M x (15.3 mL / 1000 mL/L) [Note: Convert the volume from mL to L]
Now, we can find the moles of HCl neutralized by using the balanced equation for the reaction between NaOH and HCl:
NaOH + HCl -> NaCl + H2O
For every 1 mole of NaOH, 1 mole of HCl is neutralized.
moles of HCl neutralized = moles of NaOH
b) To find the total amount of HCl needed to neutralize the antacid, we need to consider the amount required to neutralize the excess acid and the amount reacted with the CaCO3.
Total moles of HCl used = moles of HCl neutralized by NaOH + moles of HCl reacted with CaCO3
c) To calculate the mass of CaCO3 that reacted with HCl, we can find the moles of CaCO3 reacted and then convert it to grams using the molar mass of CaCO3.
We know that 1 mole of CaCO3 reacts with 2 moles of HCl according to the balanced equation:
CaCO3 + 2HCl -> CaCl2 + H2O + CO2
We have already calculated the moles of HCl reacted (from part b). So, the moles of CaCO3 reacted can be calculated by dividing the moles of HCl by 2.
moles of CaCO3 reacted = moles of HCl / 2
Mass of CaCO3 reacted = moles of CaCO3 reacted x molar mass of CaCO3
d) To find the mass percent of CaCO3 present in the antacid, we need to calculate the mass percent using the mass of CaCO3 reacted and the mass of the antacid tablet.
Mass percent of CaCO3 = (mass of CaCO3 / mass of antacid tablet) x 100%
Now, let's calculate the answers:
Given data:
Mass of antacid tablet = 5.309 g
Volume of HCl used = 30.0 mL
Concentration of HCl = 0.831 M
Volume of NaOH used = 15.3 mL
Concentration of NaOH = 0.7034 M
Molar mass of CaCO3 = 100.09 g/mol
a) Moles of NaOH used:
moles of NaOH = 0.7034 M x (15.3 mL / 1000 mL/L) = (0.7034 * 15.3) / 1000 = ?? moles
b) Total moles of HCl used = moles of HCl neutralized by NaOH + moles of HCl reacted with CaCO3
c) Moles of CaCO3 reacted:
moles of CaCO3 reacted = moles of HCl / 2
Mass of CaCO3 reacted = moles of CaCO3 reacted x molar mass of CaCO3
d) Mass percent of CaCO3 = (mass of CaCO3 / mass of antacid tablet) x 100%
The final numerical values for these variables will depend on the actual calculations. Let me know if you need further assistance or the exact values of these variables.