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.