Question

In our bodies, glucose is broken down into carbon dioxide and water, much like a combustion reaction. How many grams of O2 are needed to fully react with 150g glucose? C6H12O6(s) + 6O2(g) -->6CO2(g) + 6H2O(g)

Answer 1

Answer:

160 g

Explanation:

The chemical equation is:

C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(g)

According to the equation, 1 mol of glucose (C₆H₁₂O₆) reacts with 6 moles of O₂. We calculate the masses of the reactants from the molar masses of the chemical elements:

1 mol C₆H₁₂O₆ = (6 x 12 g/mol)+ (12 x 1 g/mol) + (6 x 16 g/mol) = 180 g

6 mol O₂ = 6 x (2 x 16 g/mol) = 6 x 32 g/mol = 192 g

So, 180 g of C₆H₁₂O₆ reacts with 192 g of O₂. The stoichiometric ratio is 192 g O₂/180 g C₆H₁₂O₆. To calculate the grams of O₂ needed to react with 150 g of C₆H₁₂O₆ we can simply multiply the stoichiometric ratio by the grams of C₆H₁₂O₆:

150 g C₆H₁₂O₆ x 192 g O₂/180 g C₆H₁₂O₆ = 160 g O₂

Therefore, 160 grams of O₂ are needed to fully react with 150 g of glucose.