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If there were 2.38 x 102 g of H2O and 18.6 moles of CO2, how much C6H12O6 can be produced?

User Scorpian
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Answer:

To solve this problem, we need to use stoichiometry, which is a way to relate the amounts of reactants and products in a chemical reaction.

The balanced equation for the reaction of glucose (C6H12O6) with oxygen (O2) to form carbon dioxide (CO2) and water (H2O) is:

C6H12O6 + 6O2 → 6CO2 + 6H2O

From the equation, we can see that for every 1 mole of glucose, 6 moles of CO2 are produced. Therefore, we can calculate the amount of glucose produced from the given amount of CO2 as follows:

18.6 moles CO2 x (1 mole C6H12O6 / 6 moles CO2) = 3.10 moles C6H12O6

Now, we need to convert the amount of H2O given in grams to moles, using the molar mass of water:

2.38 x 102 g H2O x (1 mole H2O / 18.015 g H2O) = 13.2 moles H2O

From the equation, we can see that for every 1 mole of glucose, 6 moles of H2O are produced. Therefore, the amount of glucose produced from the given amount of H2O is:

13.2 moles H2O x (1 mole C6H12O6 / 6 moles H2O) = 2.20 moles C6H12O6

Therefore, the maximum amount of glucose that can be produced is 3.10 moles or 2.20 moles, depending on which reactant limits the reaction.

Step-by-step explanation:

User Will Bolden
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