Question

For the following reaction, 13.1 grams of glucose (C6H12O6) are allowed to react with 10.6 grams of oxygen gas. glucose (C6H12O6) (s) + oxygen (g) carbon dioxide (g) + water (l) What is the maximum amount of carbon dioxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

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

Step 2: The balanced equation

C6H12O6 + 6 O2 → 6H2O + 6CO2

Step 3: 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