---
title: "A student determines the mass percent of \\(\\text{CaCO}_3\\text{(s)}\\) (molar mass \\(100.0 \\text{ g/mol}\\)) in a \\(0.500 \\text{ g}\\) antacid tablet using a back-titration procedure. The tablet is crushed and completely dissolved in \\(50.0 \\text{ mL}\\) of \\(0.200 \\text{ M } \\text{HCl(aq)}\\) (an excess). The reaction is represented by the following equation:  \\[ \\text{CaCO}_3\\text{(s)} + 2\\,\\text{H}^+\\text{(aq)} \\rightarrow \\text{Ca}^{2+}\\text{(aq)} + \\text{H}_2\\text{O(l)} + \\text{CO}_2\\text{(g)} \\]  The resulting solution is then titrated with \\(0.100 \\text{ M } \\text{NaOH(aq)}\\) to neutralize the unreacted acid according to the following equation:  \\[ \\text{H}^+\\text{(aq)} + \\text{OH}^-\\text{(aq)} \\rightarrow \\text{H}_2\\text{O(l)} \\]  If \\(20.0 \\text{ mL}\\) of \\(0.100 \\text{ M } \\text{NaOH(aq)}\\) is required to reach the equivalence point, what is the mass percent of \\(\\text{CaCO}_3\\) in the tablet?"
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url: "https://nerd-notes.com/ubq/123683/"
date_modified: "2026-09-28T12:01:58+00:00"
---

# A student determines the mass percent of \(\text{CaCO}_3\text{(s)}\) (molar mass \(100.0 \text{ g/mol}\)) in a \(0.500 \text{ g}\) antacid tablet using a back-titration procedure. The tablet is crushed and completely dissolved in \(50.0 \text{ mL}\) of \(0.200 \text{ M } \text{HCl(aq)}\) (an excess). The reaction is represented by the following equation:

\[ \text{CaCO}_3\text{(s)} + 2\,\text{H}^+\text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)} \]

The resulting solution is then titrated with \(0.100 \text{ M } \text{NaOH(aq)}\) to neutralize the unreacted acid according to the following equation:

\[ \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)} \]

If \(20.0 \text{ mL}\) of \(0.100 \text{ M } \text{NaOH(aq)}\) is required to reach the equivalence point, what is the mass percent of \(\text{CaCO}_3\) in the tablet?

A student determines the mass percent of \(\text{CaCO}_3\text{(s)}\) (molar mass \(100.0 \text{ g/mol}\)) in a \(0.500 \text{ g}\) antacid tablet using a back-titration procedure. The tablet is crushed and completely dissolved in \(50.0 \text{ mL}\) of \(0.200 \text{ M } \text{HCl(aq)}\) (an excess). The reaction is represented by the following equation:

\[ \text{CaCO}_3\text{(s)} + 2\,\text{H}^+\text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)} \]

The resulting solution is then titrated with \(0.100 \text{ M } \text{NaOH(aq)}\) to neutralize the unreacted acid according to the following equation:

\[ \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)} \]

If \(20.0 \text{ mL}\) of \(0.100 \text{ M } \text{NaOH(aq)}\) is required to reach the equivalence point, what is the mass percent of \(\text{CaCO}_3\) in the tablet?

- **A.** \(20.0\%\)
- **B.** \(40.0\%\)
- **C.** \(60.0\%\)
- **D.** \(80.0\%\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123683/*