Question

For the following reaction, 9.30 grams of glucose (C6H12O6) are allowed to react with 13.8 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:

13.7 g of CO₂

Limiting reactant:  C₆H₁₂O₆

3.81 g of O₂

Explanation:

We convert the mass of the reactants to moles, in order to find out the limiting reactant and the excess reagent

9.30 g / 180 g/mol = 0.052 moles of glucose

13.8 g / 32 g/mol = 0.431 moles of oxygen

The equation is:  C₆H₁₂O₆(s) + 6O₂ (g) → 6CO₂ (g) + 6H₂O (l)

Ratio is 1:6. Let's consider this rule of three:

1 mol of glucose reacts with 6 moles of oxygen

Then, 0.052 moles of glucose must react with (0.052 . 6) /1 = 0.312 moles

We have 0.431 moles of oxygen and we only need 0.312 moles. This means that an amount of oxygen still remains after the reaction is complete:

0.431 - 0.312 = 0.119 moles. We convert the moles to mass:

0.119 mol . 32 g / 1mol = 3.81 g

In conclussion, the limiting reactant is the glucose.

6 moles of oxygen react with 1 mol of glucose

0.431 moles of O₂ will react with (0.431 . 1) /6 = 0.072 moles of glucose

We only have 0.052 moles, so it is ok to say, that glucose is the limiting cause we do not have enough glucose.

Let's verify, the maximum amount of carbon dioxide that can be formed:

1 mol of glucose can produce 6 moles of CO₂

Therefore 0.052 moles of glucose will produce (0.052 . 6) /1 = 0.312 moles

We convert the moles to mass → 0.312 mol . 44 g /1 mol = 13.7 g