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How many grams of glucose (C6H12O6) are produced if 17.3 mol of H2O are reacted according to this balanced chemical equation?

6CO2 + 6H2O → C6H12O6 + 6O2

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Final answer:

By using the mole ratio from the balanced chemical equation and the molar mass of glucose, we can calculate that 519.48 grams of glucose are produced from 17.3 moles of water.

Step-by-step explanation:

To calculate the mass of glucose (C6H12O6) produced from the reaction of 17.3 mol of H2O, we first need to understand the balanced chemical equation:

6CO2 + 6H2O → C6H12O6 + 6O2

This equation informs us that 6 moles of H2O produce 1 mole of glucose. Therefore, we use the mole ratio to determine how many moles of glucose are produced from 17.3 mol of H2O:

(17.3 moles H2O) × (1 mole glucose / 6 moles H2O) = 2.8833 moles glucose

Now, we can find the mass of glucose produced using its molar mass:

2.8833 moles glucose × 180.156 g/mole = 519.48 grams glucose]

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