1)A typical commercial antacid tablet contains 0.350 g of CaCO3 [MW=100.09 g/mol] as an active ingredient....

1)A typical commercial antacid tablet contains 0.350 g of CaCO3 [MW=100.09 g/mol] as an active ingredient. How many moles of HCl could this tablet neutralize? 2)A student was analyzing an antacid tablet with a mass of 2.745 g. The student found that a 1.872 g sample of the tablet would neutralize 67.35 mL of stomach acid. Calculate how much stomach acid would be neutralized by the entire tablet. 3)A student determined that 29.43 mL of 0.132 M NaOH reacts with 25.0 mL of stomach acid. He dissolved a commercial tablet of antacid into 180 ml of stomach acid and then removed a 20.0 mL of the resulting solution. The Titration of this 20 mL sample took 9.56 mL of NaOH. Calculate volume of acid neutralized by antacid in the 200mL sample. 4) A 0.248 g sample of antacid containing an unknown amount of triprotic base Al(OH)3 [MW=78.00 g/mol] was reacted with 25.0 mL of 0.121 M HCl. The resulting solution was then titrated with 12.26 mL of 0.132M NaOH solution. Calculate the mass percent of Al(OH)3 in the antacid sample

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Expert Solution

1)A typical commercial antacid tablet contains 0.350 g of CaCO3 [MW=100.09 g/mol] as an active ingredient. How many moles of HCl could this tablet neutralize?

Ans: Number of moles of CaCO3 are

1 mole of CaCO3 will neutralise 2 moles of HCl. So 0.003497 moles of CaCO3 will neutralise of HCl

2)A student was analyzing an antacid tablet with a mass of 2.745 g. The student found that a 1.872 g sample of the tablet would neutralize 67.35 mL of stomach acid. Calculate how much stomach acid would be neutralized by the entire tablet.

Ans:

Hence, 98.76 mL stomach acid would be neutralized by the entire tablet.

3)A student determined that 29.43 mL of 0.132 M NaOH reacts with 25.0 mL of stomach acid. He dissolved a commercial tablet of antacid into 180 ml of stomach acid and then removed a 20.0 mL of the resulting solution. The Titration of this 20 mL sample took 9.56 mL of NaOH. Calculate volume of acid neutralized by antacid in the 200mL sample.

Ans:

25.0 mL stomach acid = 29.43 mL of NaOH

20 mL stomach acid mL of NaOH
The Titration of this 20 mL sample took 9.56 mL of NaOH.

Excess NaOH = 23.54 - 9.56 = 13.98 ml
25.0 mL stomach acid = 29.43 mL of NaOH
13.98 ml NaOH mL stomach acid
volume of acid neutralized by antacid in the 200 mL sample is 11.9 mL.
volume of acid neutralized by antacid in the 20 mL sample is mL

queen_honey_blossom answered 1 year ago

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