limiting reagents?

Question

plants use CO2, and H2O in Photosynthisis to make glucose (C6 H12 O6). if a plant has 88 grams of CO2, and 64 grams of H2O determine the following:

limiting reactant


excess reagent and mass in excess


mass of glucose produced






this one realy stumped me, I got 89.318 grams of glucose produced, and i cant figure out which is the limiting reactant and excess reagent.

thanks to whoever answers

Answer 1

the balance equation were going to write is:

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

now change 88g CO2 to moles: 88/44=2mol CO2.
64g H2O to moles: 64/18= 3.55mol H2O

to find the limiting factor(reagant, i like calling it factor) is CO2 because 2/6 is less than 3.55/6. now make a chart.(ill draw the chart and explain after)

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

mole........2.............3.55...............0..................0
change...-2.............-2................+033.............+2
final..........0............1.55..............0.33...............2

the limiting factor went down all the way to 0. the H20 went down the same amount as CO2, because they have the same coefficient. glucose started out at 0, because we assume there is no product befor the reaction. It goes up 1/6 of what CO2 went down(the ratio of the coefficients)

now: they asked for whats left. now change the moles into grams again by multiplying by the molar mass.

1.55 mol H2Ox18= 27.9g

0.33moles of C6H12O6 x 180 =59.4 g

so ur answers:

Limiting factor is CO2

excess factor is H2O and ur left with 27.9g

u formed 59.4g of C6H12O6

if u have any questions, feel free to contact me.

Answer 2

OK..so you've got the balanced equation as:
6 CO2 + 6 H2O --> C6H12O6 + 6 O2

Before you can determine the mass of glucose formed, you HAVE to determine the limiting reactant. So, you basically have to solve the problem twice, starting with your two different reactants.

So, starting with 88 grams of CO2, convert that to moles using the molar mass of CO2. Next, from the balanced equation, you see that you can make 1 mole of glucose for every 6 moles of CO2. Finally convert the moles of glucose to grams of glucose using its molar mass.

Now, repeat that calculation, but this time start with the mass of water. Convert that to moles, then use the coefficients of the balanced equation to convert moles of water to moles of glucose, and finally convert moles of glucose to grams.

The reactant which produces the smaller mass of glucose is your limiting reactant, and you can produce that many grams of glucose, and no more. I calculated that you can form 60.0 grams of glucose.

By my calculations, CO2 is the limiting reactant. Now, you need to find out the moles and mass of the water that are left. Since water and CO2 are used in a 1:1 mole ratio in this reaction, the you can calculate the moles of water you started with and subtract the moles of CO2 that you started with. The difference is the moles of water remaining. Multiply that by the molar mass of water to get the mass of water remaining.

Hope this helps...

Answer 3

Look at the number of moles you need for each compound individually in light of the molecular weight of the compound.
So let's say you have 88 g of CO2, look at the molecular weight of CO2 (in g/mol) and divide 88 g by that number to obtain the # of mol of CO2. Do the same for H2O and then compare in light of the formula; which one has the smallest amount?

They use CO2 and H2O in equal quantities to make glucose (i.e. 6 CO2 + 6 H2O --> C6H12O6 + 6O2

I'm not gonna work out the problem for you (I would, but I gotta go) but good luck! :)