Question

Experiment 12 THE DETERMINATION OF THE NEUTRALIZING ABILITY OF ANTACIDS The Pre-Lab 1. A 1.032 gram sample of an antacid was dissolved in 50 mL of water. Then a 49.33 mL portion of 0.488 M HCI was added to the flask. The antacid was completely neutralized and the unneutralized hydrochloric acid remaining in the flask required 29.33 mL of 0.510 M NaOH to completely neutralize it. a) The source of all of the acid used in the experiment is the hydrochloric acid, HCI. What are the sources of the two bases that react with this HCI b) How many moles of the acid, HCI, are used in the experiment? moles of solute = volume in liten X concentration solute (in moles/liter) c) What must be the total of the number of moles of HCl neutralized by the antacid plus the number of moles of HCl neutralized by the NaOH? d) Calculate the number of moles of the NaOH added during the experiment. moles of solute = volume (in liters) X concentration solute (In moleiter) e) How many moles of the total moles of HCl originally present does the added NaOH react with during its neutralization reaction? See Figure 12.2a 1) How many moles of the total moles of HCl originally present must the antacid react with during its neutralization reaction? g) How many moles of HCl are neutralized for each gram of antacid tablet used?
this is my prelab for class, I'm really struggling to understand so if someone could help me break down the problems that would be nice

Answer 1

Information provided is listed as:

Antacid taken = 1.032g in 50 mL water

HCl added = 49.33 mL of 0.488 M concentration

After neutralizing the antacid, NaOH required to neutralize the remained HCl = 29.33 mL of 0.510 M NaOH

a). The source of the two bases that react with HCl is antacid and Sodium hydroxide.

b). No. of moles of acid used in experiment = concentration (M) * volume (in L)

= 0.488 M * (49.33/1000) = 0.024 moles

c). total no. of moles of HCl that must be neutralized by antacid and NaOH is 0.024 moles

d). No. of moles of NaOH added = concentration (M) * volume (in L)

= 0.510* (29.33/1000) = 0.015 moles