The decomposition reaction of ${\text{KClO}}_{3}$(s) is catalyzed by ${\text{MnO}}_{2}$(s). The reaction produces ${\text{KCl}}$(s) and ${\text{O}}_{2}$(g). In one experiment 12.24 g of ${\text{KClO}}_{3}$(s) and 0.87 g of ${\text{MnO}}_{2}$(s) are used, and 9.60 g of ${\text{O}}_{2}$(g) are produced. In second experiment, the mass of ${\text{KClO}}_{3}$(s) is halved and the mass of ${\text{MnO}}_{2}$(s) is kept the same (0.87 g). In the second experiment, the mass of ${\text{O}}_{2}$(g) produced is one-half that in the first experiment. In a third experiment, 12.24 g of ${\text{KClO}}_{3}$(s) and one-half the grams of ${\text{MnO}}_{2}$(s) are used, how does the mass of ${\text{O}}_{2}$(g) produced in the third experiment compare to the mass of ${\text{O}}_{2}$(g) in the first experiment? Molar masses (in g mol${}^{-1}$) ${\text{KClO}}_{3}=122.4$, ${\text{MnO}}_{2}=86.94$, ${\text{O}}_{2}=32.00$, ${\text{KCl}}=74.55$

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