To determine how many moles of [tex]\( \text{CO}_2 \)[/tex] are produced when 6.0 moles of [tex]\( \text{O}_2 \)[/tex] are used, we will refer to the balanced chemical equation provided:
[tex]\[ \text{C}_6\text{H}_{12}\text{O}_6 + 6 \text{O}_2 \rightarrow 6 \text{CO}_2 + 6 \text{H}_2\text{O} \][/tex]
This equation tells us that one mole of glucose ([tex]\( \text{C}_6\text{H}_{12}\text{O}_6 \)[/tex]) reacts with six moles of oxygen gas ([tex]\( \text{O}_2 \)[/tex]) to produce six moles of carbon dioxide ([tex]\( \text{CO}_2 \)[/tex]) and six moles of water ([tex]\( \text{H}_2\text{O} \)[/tex]).
The key aspect here is the stoichiometric relationship between [tex]\( \text{O}_2 \)[/tex] and [tex]\( \text{CO}_2 \)[/tex] from the balanced equation. The coefficients in the balanced equation show that:
[tex]\[ 6 \text{ moles of } \text{O}_2 \rightarrow 6 \text{ moles of } \text{CO}_2 \][/tex]
So, if we start with 6.0 moles of [tex]\( \text{O}_2 \)[/tex]:
[tex]\[ 6.0 \text{ moles of } \text{O}_2 \text{ will produce } 6.0 \text{ moles of } \text{CO}_2 \][/tex]
Therefore, the number of moles of [tex]\( \text{CO}_2 \)[/tex] produced when 6.0 moles of [tex]\( \text{O}_2 \)[/tex] are used is:
[tex]\[ 6.0 \text{ moles of } \text{CO}_2 \][/tex]
