Final answer:
To determine the limiting reactant for the reaction between sucrose and oxygen, moles of each reactant were calculated. The stoichiometry of the balanced equation indicates that sucrose is the limiting reactant because there isn't enough oxygen to fully react with the available sucrose.
Step-by-step explanation:
The student asked which is the limiting reactant when 10 g sucrose (C12H22O11) and 10 g oxygen (O2) are reacting according to the following balanced equation:
C12H22O11 + 12O2 → 12CO2 + 11H2O.
Step 1: Identify the molar mass of the reactants. Sucrose (C12H22O11) has a molar mass of about 342.3 g/mol. Oxygen (O2) has a molar mass of about 32.0 g/mol.
Step 2: Convert the masses of the reactants to moles. For sucrose, 10 g divided by 342.3 g/mol equals about 0.0292 mol. For oxygen, 10 g divided by 32.0 g/mol equals about 0.3125 mol.
Step 3: Use the stoichiometry of the reaction. For each mole of sucrose, 12 moles of oxygen are required. Multiply the number of moles of each reactant by its coefficient in the balanced equation.
Step 4: The reactant that produces a smaller amount of product is the limiting reactant. Since we only have 0.0292 mol of sucrose, and it would need 0.3504 mol of oxygen (0.0292 mol sucrose × 12 mol O2/mol sucrose), but we only have 0.3125 mol of oxygen available, oxygen is the excess reactant. Therefore, sucrose is the limiting reactant for this reaction.