Answer:
[tex]m_{O_2}=61.87gO_2[/tex]
[tex]m_{CO_2}=85.07gCO_2[/tex]
Explanation:
Hello,
Considering the given reaction's stoichiometry, grams of oxygen result:
[tex]m_{O_2}=58.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molO_2}{1molC_6H_{12}O_6}*\frac{32gO_2}{1molO_2}\\m_{O_2}=61.87gO_2[/tex]
Moreover, the mass of produced carbon dioxide turns out:
[tex]m_{CO_2}=58.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molCO_2}{1molC_6H_{12}O_6}*\frac{44gCO_2}{1molCO_2}\\m_{O_2}=85.07gCO_2[/tex]
Best regards.