In the combustion analysis of 0.1127 g of glucose (C₆H₁₂O₆), what mass, in grams, of CO₂ would be produced? A)0.0451 g B)0.0825 g C)0.1652 g D)0.4132 g E)1.466 g

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Quizplus · Accepted Answer

The molecular formula for glucose is C6H12O6. The combustion of glucose can be represented by the balanced chemical equation: C6H12O6 + 6O2 → 6CO2 + 6H2O. This means 1 mole of glucose produces 6 moles of CO2. The molar mass of glucose is approximately 180.16 g/mol, and the molar mass of CO2 is approximately 44.01 g/mol. Given 0.1127 g of glucose, the moles of glucose are 0.1127 g / 180.16 g/mol = 0.0006257 mol. Therefore, the moles of CO2 produced are 6 times this amount, which is 0.0037542 mol. The mass of CO2 produced is 0.0037542 mol * 44.01 g/mol = 0.1652 g.

In the Combustion Analysis of 0