To determine how many moles of [tex]\( \text{CO}_2 \)[/tex] are produced from the fermentation of a given amount of glucose ([tex]\( \text{C}_6 \text{H}_{12} \text{O}_6 \)[/tex]), we need to look at the balanced chemical equation for the reaction:
[tex]\[
\text{C}_6 \text{H}_{12} \text{O}_6( aq ) \rightarrow 2 \text{C}_2 \text{H}_5 \text{OH} ( aq )+2 \text{CO}_2(g)
\][/tex]
This equation tells us that one mole of glucose ([tex]\( \text{C}_6 \text{H}_{12} \text{O}_6 \)[/tex]) produces two moles of carbon dioxide ([tex]\( \text{CO}_2 \)[/tex]).
Given:
- Moles of glucose ([tex]\( \text{C}_6 \text{H}_{12} \text{O}_6 \)[/tex]) reacting = 0.300 mol
We can use the stoichiometric coefficients from the balanced equation to convert the moles of glucose to moles of carbon dioxide. According to the balanced equation, the ratio of moles of glucose to moles of carbon dioxide is 1:2. This means for every 1 mole of glucose, 2 moles of carbon dioxide are produced.
Step-by-step solution:
1. Identify the moles of glucose reacting: 0.300 mol
2. Use the stoichiometric ratio from the balanced equation:
[tex]\[
1 \text{ mol } \text{C}_6 \text{H}_{12} \text{O}_6 : 2 \text{ mol } \text{CO}_2
\][/tex]
3. Calculate the moles of [tex]\( \text{CO}_2 \)[/tex] produced:
[tex]\[
\text{Moles of } \text{CO}_2 = 0.300 \text{ mol of } \text{C}_6 \text{H}_{12} \text{O}_6 \times \frac{2 \text{ mol } \text{CO}_2}{1 \text{ mol } \text{C}_6 \text{H}_{12} \text{O}_6}
\][/tex]
[tex]\[
\text{Moles of } \text{CO}_2 = 0.300 \times 2 = 0.600 \text{ mol}
\][/tex]
Thus, when 0.300 mol of [tex]\( \text{C}_6 \text{H}_{12} \text{O}_6 \)[/tex] reacts, 0.600 mol of [tex]\( \text{CO}_2 \)[/tex] is produced.
