Question

Glucose, C6H12O6, is used as an energy source by the human body. The overall reaction in the body is described by the equation:

Answer 1

Answer:

1) $m_{O_2}=24.5gO_2$

2) $m_{CO_2}=33.7gCO_2$

Explanation:

Hello,

1) In this case, by considering the given reaction, starting by 23.0 g of glucose, the grams of oxygen by stoichiometry result:

$m_{O_2}=23.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molO_2}{1molC_6H_{12}O_6}*\frac{32gO_2}{1molO_2} \\m_{O_2}=24.5gO_2$

2), Now, by means of also stoichiometry, the grams of carbon dioxide that are produced result:

$m_{CO_2}=23.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molCO_2}{1molC_6H_{12}O_6}*\frac{44gCO_2}{1molCO_2}\\m_{CO_2}=33.7gCO_2$

Best regards.