A sample of limestone and other soil materials was heated, and the limestone decomposed to give calcium oxide and carbon dioxide. CACO3 (s)→ Ca0(s) + CO2(g) A 4.138 g sample of limestone-containing material gave 1.49 g of CO2, in addition to CaO, after being heated at a high temperature. What was the mass percent of CACO3 in the original sample? 63.9 %

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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**Understanding Chemical Reactions: Decomposition of Limestone**

*Chemical Reaction Overview:*
A sample of limestone and other soil materials was heated, resulting in the decomposition of limestone to produce calcium oxide (CaO) and carbon dioxide (CO₂).

The chemical equation for this reaction is:
\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \]

*Problem Statement:*
A 4.138 g sample of limestone-containing material yielded 1.49 g of CO₂ after being heated at high temperature. Determine the mass percent of CaCO₃ in the original sample.

*Calculation:*
A value box is provided with the answer of 63.9%.

Please ensure you understand the fundamentals of stoichiometry and mass percentage calculations to solve similar problems.
Transcribed Image Text:**Understanding Chemical Reactions: Decomposition of Limestone** *Chemical Reaction Overview:* A sample of limestone and other soil materials was heated, resulting in the decomposition of limestone to produce calcium oxide (CaO) and carbon dioxide (CO₂). The chemical equation for this reaction is: \[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \] *Problem Statement:* A 4.138 g sample of limestone-containing material yielded 1.49 g of CO₂ after being heated at high temperature. Determine the mass percent of CaCO₃ in the original sample. *Calculation:* A value box is provided with the answer of 63.9%. Please ensure you understand the fundamentals of stoichiometry and mass percentage calculations to solve similar problems.
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