Answer:
There will be formed 14.58 grams of CO2
O2 is the limiting reagent
There will remain 3.151 grams of glucose
Explanation:
Step 1: Data given
Mass of glucose = 13.1 grams
molar mass of glucose 180.156 g/mol
Mass of oxygen = 10.6grams
molar mass of oxygen = 32 g/mol
<u>Step 2</u>: The balanced equation
C6H12O6 + 6 O2 → 6H2O + 6CO2
<u>Step 3</u>: Calculate moles of glucose
Moles of glucose = mass glucose / molar mass glucose
Moles of glucose = 13.1 grams / 180.156 g/mol
Moles of glucose = 0.0727 moles
Step 4: Calculate moles of oxygen
Moles O2 = 10.6 grams / 32 g/mol
Moles O2 = 0.33125 moles
Step 5: Calculate the limiting reactant
For 1 mol of glucose , we need 6 moles of O2 to produce 6 moles of H2O and 6 moles of CO2
Oxygen is the limiting reactant. It will completely be consumed ( 0.33125 moles).
Glucose is in excess. There will be consumed 0.33125/6 = 0.05521 moles
There will remain 0.0727 - 0.05521 = 0.01749 moles
This is 0.01749 moles * 180.156 g/mol = 3.151 grams
Step 6: Calculate moles of CO2
For 1 mol of glucose , we need 6 moles of O2 to produce 6 moles of H2O and 6 moles of CO2
For 0.33125 moles of O2 we'll get 0.33125 moles of CO2 produced
Step 7: Calculate mass of CO2
Mass of CO2 = moles CO2 * molar mass CO2
Mass CO2 = 0.33125 moles * 44.01 g/mol
Mass CO2 = 14.58 grams
