A student made the following diagram to represent cellular respiration.

The title of the diagram is Cellular Respiration. An oval labeled Cell has five arrows around it. The arrow labeled Oxygen points towards the cell. The other four arrows labeled Glucose, Carbon dioxide, Water, and Energy point away from the cell.

Why is this diagram incorrect?

Water is not formed during this process.
Carbon dioxide is not formed during this process.
The arrow for oxygen should point in the opposite direction.
The arrow for glucose should point in the opposite direction.

Hi there! So this is for your Science finals I see. I take the previous answer back as I just realized that's incorrect. During cellular respiration, carbon dioxide (CO2) is formed when oxygen combines with the carbon atoms from glucose. It is an ending compound.

Water does end up getting formed, since glucose reacts with oxygen, forming ATP that can be used by the cell. Carbon dioxide and water are created as byproducts. ... Water and carbon dioxide are released as byproducts.

^ that is correct, but we're looking for why the diagram is INCORRECT.

Carbon Dioxide is formed during the process and it is a waste byproduct.

Oxygen is pointing TO the cell. This is also correct! Cellular Respiration is practically the opposite of Photosynthesis. It takes in oxygen, and releases CO2. Photosynthesis takes in CO2 and releases oxygen.

Glucose is a starting compound, and it points away from the cell. This is actually, incorrect. As glucose breaks down it creates energy. So, this is your answer.

If you read it this far, congrats. Your answer is  D, The arrow for glucose should point in the opposite direction.

Glucose is stored to power the rest of the cell.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g