In the complete combustion of glucose (C6H12O6), how many moles of C6H12O6(s) are required to produce 24 moles of CO2(g)?

• A. 4 moles
• B. 24 moles
• C. 12 moles
• D. 6 moles

Answer: (A)

Verified step by step guidance

1. Write the balanced chemical equation for the complete combustion of glucose (C6H12O6). The general form is: \(\mathrm{C_6H_{12}O_6 (s) + O_2 (g) \rightarrow CO_2 (g) + H_2O (g)}\).
2. Balance the carbon atoms first. Since glucose has 6 carbon atoms, it will produce 6 moles of \(\mathrm{CO_2}\) per mole of glucose. So, the balanced equation starts as: \(\mathrm{C_6H_{12}O_6 (s) + O_2 (g) \rightarrow 6 CO_2 (g) + H_2O (g)}\).
3. Balance the hydrogen atoms next. Glucose has 12 hydrogen atoms, which form water molecules. Since each water molecule has 2 hydrogens, glucose produces 6 moles of \(\mathrm{H_2O}\). The equation now is: \(\mathrm{C_6H_{12}O_6 (s) + O_2 (g) \rightarrow 6 CO_2 (g) + 6 H_2O (g)}\).
4. Balance the oxygen atoms last. On the right side, there are \$6 \times 2 = 12\( oxygens from \)\mathrm{CO_2}\( and \)6 \times 1 = 6\( oxygens from \)\mathrm{H_2O}\(, totaling 18 oxygen atoms. Since glucose provides 6 oxygens, the remaining 12 oxygens must come from \)\mathrm{O_2}\(. Because each \)\mathrm{O_2}\( molecule has 2 oxygen atoms, you need 6 moles of \)\mathrm{O_2}\(. The balanced equation is: \)\mathrm{C_6H_{12}O_6 (s) + 6 O_2 (g) \rightarrow 6 CO_2 (g) + 6 H_2O (g)}$.
5. Use the mole ratio from the balanced equation to find how many moles of glucose are needed to produce 24 moles of \(\mathrm{CO_2}\). Since 1 mole of glucose produces 6 moles of \(\mathrm{CO_2}\), set up the proportion: \(\frac{1 \text{ mole glucose}}{6 \text{ moles } CO_2} = \frac{x \text{ moles glucose}}{24 \text{ moles } CO_2}\). Solve for \(x\) to find the required moles of glucose.