Question

For the following reaction, 16.0 grams of glucose (C6H12O6) are allowed to react with 11.9 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

We will deal the problem in terms of moles.

C₆H₁₂O₆   + 6O₂   -------> 6CO₂   + 6H₂O

According to stoichiometric coefficient of equation 1 mole of glucose reacts with 6 mole of oxygen to get 6 mole of carbon dioxide and 6 mole of water.

Molar mass of C₆H₁₂O₆ = 12*6 + 1*12 + 16*6

= 180 g/mol

Given mass of glucose = 16g

Moles = Given mass/Molar mass

Moles of glucose = 16/180 =0.088

Molar mass of O₂ = 16*2 = 32g/mol

Given mass of O₂ = 11.9g

Moles of O₂ = 11.9/32 = 0.37

1 mole of glucose reacts with 6 moles of oxygen.

Moles of glucose that would react with 1 mole of oxygen = 1/6

Moles of glucose that would react with 0.37 moles of oxygen = 1/6*(0.37) =0.061

But we have 0.088 mole of glucose available , therefore glucose is present in excess.

6 mole of O₂ gives 6 mole of carbon dioxide .

Moles of CO₂ obtained from 0.37 moles of O₂ = 0.37

Mass of CO₂ contained in 0.37 moles = 0.37*44 =16.28g

Limiting reagent is the substance which is completely reacted in a reaction.

Clearly oxygen is completely reacted , so it is a limiting reagent.

Total glucose available = 0.088

Glucose reacted = 0.061

Glucose unreacted = 0.088 - 0.061 = 0.027

Mass of glucose unreacted = 0.027*180 = 4.86g

Maximum amount of CO₂ formed = 16.26g

Formula of limiting reagent = O₂

Amount of excess reagent left = 4.86g (glucose)