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I'm working on a question which asks me to do two things. I've balanced the related equation as:

$$\ce{6H2O + 6CO2 -> C6H12O6 + 6O2}$$

Now, if I have $44$ grams of $\ce{CO2}$ and $36$ grams of $\ce{H2O}$, how many moles of glucose will I have? The molecular mass of $\ce{CO2}$ is $44\mathrm{~g/mole}$, and that of $\ce{H2O}$ is $18\mathrm{~g/mole}$. Do I need to determine how many moles of each element are in the certain amount of grams for each compound, and then use that to determine how many moles are in the glucose?

Secondly, for the same amounts of $\ce{CO2}$ and $\ce{H2O}$, I need to figure the theoretical yield. Am I correct in saying since for ever 6 molecules of $\ce{H2O}$ and $\ce{CO2}$ I get 1 molecule of glucose, I would do

$$36\text{ grams of } \ce{H2O} \times \left( \frac{1~\mathrm{mol}~\ce{H2O}}{18\mathrm{g}~\ce{H2O}} \right) \times \left(\frac{1~\mathrm{mol}~\ce{C6H12O6}}{6~\mathrm{mol}~\ce{H2O}}\right)$$

Then take the same for $\ce{CO2}$? I'm unsure of what to do with the values after that.

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  • $\begingroup$ @IͶΔ point noted. $\endgroup$ – Aditya Dev Mar 17 '16 at 15:03
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You are correct in your thinking.

First, knowing the mass of the compounds, use the molar mass relationship for each (g/mol) to convert them all to moles. This is because when it comes to reactions, compounds react on a mole or number basis. This lets you know the "starting amounts" of $\ce{H_2O}$ and $\ce{CO_2}$. Now, our forum has a policy against providing direct answers to homework, so I'll give examples. If you hypothetically had, say, 12 moles of both water and carbon dioxide, then using the numbers in the balanced chemical equation, they can fully react to make 2 moles of glucose. Thus, 2 moles is your theoretical yield. However, in another case, if you had 9 moles of water and 14 moles of carbon dioxide, the $\ce{H_2O}$ limits the amount of glucose you can produce, and it is the limiting reactant. (14 moles of $\ce{CO_2}$ would require 14 moles of $\ce{H_2O}$. There will be some $\ce{CO_2}$ left over because you are limited by only 9 moles of $\ce{H_2O}$). In this hypothetical case, the theoretical yield would be 1.5 moles of glucose. (Your problem may or may not involve limiting/excess reactants).

And if you're wondering, it's called "theoretical" yield because if you were carrying out this reaction in real life, some inefficiencies in combustion would yield less than the calculated amount. Hope this helps.

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  • $\begingroup$ This is immensely helpful, thank you! So since 44g of CO2 is equal to one mole, and 36g H2O is equal to 2 moles, and I need 6 moles of each for a proper reaction without any left over, then the theoretical yield would be 1/6 a mole of glucose correct? $\endgroup$ – Paul Mar 17 '16 at 2:23
  • $\begingroup$ It's not that you need a minimum of 6 moles of each to make a complete reaction. What the equation tells you is that carbon dioxide and water react in a 1:1 ratio to form 1/6 the amount of glucose. Since you have 2 moles of water and only 1 mole of carbon dioxide, your amount of carbon dioxide determines the maximum amount of glucose you can make. And like you said, 1 mole of carbon dioxide is stoichiometrically equal to 1/6 mole of glucose. $\endgroup$ – Yunfei Ma Mar 17 '16 at 15:11

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