Question

hey guys..
give me an example of limiting reagent ..from fundamental concepts.....

Answer 1

EXAMPLE  1.1 : PHOTOSYNTHESIS

Consider respiration, one of the most common chemical reactions on earth.

C6H12O6+6O2→6CO2+6H2O+energy(1.1)

What mass of carbon dioxide forms in the reaction of 25 grams of glucose with 40 grams of oxygen?

SOLUTION

When approaching this problem, observe that every 1 mole of glucose ( C6H12O6 ) requires 6 moles of oxygen to obtain 6 moles of carbon dioxide and 6 moles of water.  

Step 1: Determine the balanced chemical equation for the chemical reaction.

The balanced chemical equation is already given.

Step 2: Convert all given information into moles (most likely, through the use of molar mass as a conversion factor).

25g×1mol180.06g=0.1388molC6H12O6  

40g×1mol32g=1.25molO2  

Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.

a. If all of the 1.25 moles of oxygen were to be used up, there would need to be  1.25×16  or 0.208 moles of glucose. There is only 0.1388 moles of glucose available which makes it the limiting reactant.

1.25molO2×1molC6H12O66molO2=0.208molC6H12O6

 

b. If all of the 0.1388 moles of glucose were used up, there would need to be 0.1388 x 6 or 0.8328 moles of oxygen. Because there is an excess of oxygen, the glucose amount is used to calculate the amount of the products in the reaction.

0.1388molC6H12O6×6molO21molC6H12O6=0.8328molO2

 

If more than 6 moles of O2 are available per mole of C6H12O6, the oxygen is in excess and glucose is the limiting reactant. If less than 6 moles of oxygen are available per mole of glucose, oxygen is the limiting reactant. The ratio is 6 mole oxygen per 1 mole glucose, OR 1 mole oxygen per 1/6 mole glucose. This means: 6 mol O2 / 1 mol C6H12O6 .

Therefore, the mole ratio is: (0.8328 mol O2)/(0.208 mol C6H12O6)

This gives a 4.004 ratio of O2 to C6H12O6.

Step 4: Use the amount of limiting reactant to calculate the amount of CO2 or H2O produced.

For carbon dioxide produced:  0.1388molesglucose×61=0.8328molescarbondioxide .

Step 5: If necessary, calculate how much is left in excess.

1.25 mol - 0.8328 mol = 0.4172 moles of oxygen left over