Certainly! Let's solve this step-by-step:
1. Identify the balanced chemical equation for the reaction:
[tex]\[ C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O \][/tex]
2. Look at the stoichiometry of the equation:
This equation tells us that 1 mole of glucose ([tex]\(C_6H_{12}O_6\)[/tex]) reacts with 6 moles of oxygen ([tex]\(O_2\)[/tex]) to produce 6 moles of carbon dioxide ([tex]\(CO_2\)[/tex]) and 6 moles of water ([tex]\(H_2O\)[/tex]).
3. Determine the mole ratio:
From the balanced equation, we see that 6 moles of [tex]\(O_2\)[/tex] produce 6 moles of [tex]\(CO_2\)[/tex]. Thus, the ratio of [tex]\(O_2\)[/tex] to [tex]\(CO_2\)[/tex] is 1:1.
4. Calculate the moles of [tex]\(CO_2\)[/tex] produced:
Since the ratio of [tex]\(O_2\)[/tex] to [tex]\(CO_2\)[/tex] is 1:1, for every mole of [tex]\(O_2\)[/tex] used, 1 mole of [tex]\(CO_2\)[/tex] is produced. Given that 6.0 moles of [tex]\(O_2\)[/tex] are used, this directly produces:
[tex]\[ 6.0 \text{ moles of } CO_2 \][/tex]
Therefore, if 6.0 moles of [tex]\(O_2\)[/tex] are used, 6.0 moles of [tex]\(CO_2\)[/tex] are produced.
