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Part iic. scoring scheme: 3-3-2-1. based upon the molarities of the standardized hcl and standardized naoh, and the volumes of each added to analyze the antacid samples, calculate the millimoles of hcl needed to neutralize the tablet per milligram of sample. use the same sequence as you entered the mass of the samples and report your results to 4 significant figures.

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Equivalent millimoles HCl/mg sample = ( (concentration HCl in mmol/mL)(Volume HCl in mL) - (concentration NaOH in mmol/mL)(Volume NaOH in mL) ) / ( (1000 mg/g)(mass antacid sample in grams ) 
From this equation, then you will be guided.

Moles = Molarity * liters

But if we are calculating millimoles, we may use mL.
Using this:
Moles reacted with antacid = moles of HCl - moles of NaOH
Moles reacted = 0.4548 * 14.1 - 0.1048 * 14.6
Moles reacted = 4.88 millimoles

The mass of the tablet was 1.333 grams = 1333  mg

Millimoles of acid per mg tablet = 4.88/1333

There are about 0.0037 millimoles reacting per milligram of tablet.

About 0.0037 millimoles should react per milligram of the tablet.

To calculate the millimoles of HCl needed to neutralize the tablet per milligram of the sample, the following expressions will help:

[tex]\frac{\rm{Equivalent \;millimoles\; HCl}}{\rm{mg\; sample}}= \frac{(\rm{concentration\; of\; HCl)} \times (Volume\;of\; HCl\; in\; mL - concentration\;of\; NaOH) \times Volume \;NaOH }{( (1000\;\rm mg/g)\;(mass\; antacid\; sample \;in \;grams)}}[/tex]

Moles = Molarity [tex]\times[/tex] Volume(L)

Moles reacted with antacid = Moles of HCl -moles of NaOH

= (0.4548 [tex]\times[/tex] 14.1) - (0.1048 [tex]\times[/tex] 14.6)

= 4.88 mmoles.

Mass of tablet = 1.33 grams

Mass of tablet = 1333 mg

Milimoles of acid per mg tablet = [tex]\rm \dfrac{4.88}{1333}= \rm{0.0037\; millimoles}[/tex]

Milimoles of acid per mg tablet = 0.0037 mmoles.

Thus, per milligram of tablets, about 0.0037 millimoles react.

To learn more about neutralization reaction, refer to the link:

https://brainly.com/question/15255706

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