March 19, 2012

On March 19, 2012, we discussed limiting reagents and how it relates to making a hamburger. We had to make one double hamburger with bacon and cheese and the formula was as follows: 1 Hamburger bun + 1 Hamburger Patty + 2 Slices of Cheese + 4 Slices of Bacon = 1 Quarter Pounder. We had 10 pounds of Hamburger, 5 pounds of cheese, 4 pounds of bacon and 4 dozen buns. What we are looking to do is figure out what ingredient is the limiting reagent, or which ingredient will get run out of first when making the hamburgers. My process of figuring out how many hamburgers can I make started out by solving for how many buns I have and the number of hamburgers that can go on the buns. After this, I moved on the cheese and bacon. The first piece of information I needed was how many slices of cheese I could get from 5 pounds so in order to do so I looked up how much 1 slice of cheese weighed. This helped with the conversion factors to solve for how many slices of cheese I have. Next, I performed a similar process with the bacon. I researched how many slices of bacon I could get from 1 pound. The number of slices depended on if it was thick or thin sliced. Once I knew how many slices, the conversion factor was used and I determined the maximum number of bacon slices I will have to make as many hamburgers as possible. The maximum number of burgers depended on if you were looking at the bacon or cheese as the limiting reagent. When using thin strips over thick strips of bacon, more strips could be used and therefore more burgers could be made. This also affected weather enough cheese was available as well. I calculated how many hamburgers I could make from using thin strips of bacon so therefore bacon was the limiting reagent. I do not remember the exact number of burgers I estimated but this was my processing.

For the reaction of 2 moles of hydrogen and 1 mole of oxygen resulted in 2 moles of water, I was trying to determine the which is the limiting reagent and how much product can be made beginning with 8.5 grams of hydrogen and 40.2 grams of oxygen. The following steps can be used:

1.       Write balanced equation

2.       Calculate moles of reagents

3.       Determine limiting reagent

4.       Calculate moles of products

5.       Calculate mass of products

6.       Mass Collected

Here is the work following along with the above steps.

1.       2H₂ + O₂ à 2H₂O

2.       Moles=mass/molar mass

Moles H₂= 8.5grams/(2grams/mole)= 4.25 moles H₂

Moles O₂= 40.2grams/(32grams/mole)= 1.256 moles O₂

3.       Limiting Reagent

Two (2) multiplied by 1.256= 2.513 moles O₂

The excess of H₂ means that O₂ is the limiting reagent.

2         Moles of H₂O produced (based on limiting reagent)

1.256 moles O₂ multiplied by (2 moles H₂O/1 mole O₂)= 2.513 moles H₂O

3         Mass of H₂O

4         250 moles H₂O multiplied by (18.0 grams H₂O/1 mole O₂)= 112.5 grams of H₂O

5         Amount of H₂ remaining

4.25 – 2.513 = 1.738 grams H₂

Final Answer:

112.5 grams of H₂O

Oxygen is the limiting reagent

1.738 grams of H₂ product remaining