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Chemistry question- In chemistry lab we were determining the composition of a TUMs tablet. We wanted...

Chemistry question- In chemistry lab we were determining the composition of a TUMs tablet. We wanted to find out how much calcium carbonate there was in the TUMs tablet and how much HCl was neutralized by the TUMS tablet (calcium carbonate). The questions that we are supposed to answer are A) calculate mols of NaOH used in titration B) Calculate mols of HCl neutralized by TUMs tablet C) Calculate mols of CaCO3 reacted D) Calculate grams of CaCO3 reacted E) Calculate the percentage of CaCO3 in TUMs tablet. I have found A which is 0.4978 mols NaOH. The data for this experiment are as follows: 19.913 g NaOH used to make 1 M solution mixed in 500 mL volumetric flask. 38 mL 38% HCl used to make 1 Molar solution, mass of one TUMs tablet 1.305 g, mass of TUMs tablet when crushed and used in titration 1.302, amount of 1 M HCl solution put into TUMs beaker 29.0 mL, amount of 1 M NaOH solution put into CONTROL beaker to turn phenolphthalein pink 19.1 mL, amount of 1 M NaOH put into TUMs beaker to turn phenolphthalein pink 9.19 mL. Please walk me through the steps so that I can understand how to do this. Thanks in advance and if any other information is needed just let me know and I will provide that. Oh and experiment was done at room temperature.

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Answer #1

moles of NaOH = 0.4978

mass of TUMs tablet = 1.302

9.19 mL of 1M NaOH were neutralized by HCl, these means:

0.00919 L * 1 mol/L = 0.00919 moles of NaOH neutralized, which means 0.00919 moles of HCl were consumed by NaOH

This means, that the rest were neutralized by TUMs. To get them, we get total moles of HCl in TUMs beaker:

29 mL = 0.029 L * 1mol/L = 0.029 moles of HCl

Moles of HCl neutralized by TUMs = 0.029 - 0.00919 = 0.01981 moles of HCl, answer to letter B

For letter C, we have CaCO3 reaction with HCl:

2HCl + CaCO3 -> CaCl2 + H2CO3

To get the moles of CaCO3 that reacted:

0.01981 moles of HCl * (1 mol of CaCO3 / 2 moles of HCl) = c) 0.0099 moles of CaCO3

We convert them to grams:

0.0099 moles of CaCO3 * (100 g/mol) = d) 0.9905 g of CaCO3

Percentage of tablet reacted = 0.9905 / 1.305 =e) 75.9%

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