# Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179c: 27

$84.28\%$

#### Work Step by Step

Number of moles of $CO_2$: $0.558\ g/44.001\ g/mol=0.0127\ mol$ From stoichiometry, that's the number of moles of calcium carbonate. $0.0127\ mol\times 100.087\ g/mol=1.269\ g$ Mass fraction: $1.269\ g/1.506\ g=84.28\%$

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