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6 September, 07:36

Glucose, C6H12O6, is metabolized in living systems. How many grams of water can be produced from the reaction of 18.0 g of glucose and 7.50 L of O2 at 1.00 atm and 37 °C? C6H12O6 (s) + 6O2 (g) - > 6CO2 (g) + 6H2O (l)

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  1. 6 September, 07:57
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    There is 5.32 grams of water produced

    Explanation:

    Step 1: Data given

    Mass of glucose = 18.0 grams

    Molar mass of glucose = 180.156 g/mol

    Volume of O2 = 7.50 L

    Temperature = 37 °C

    Step 2: The balanced equation

    C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l)

    Step 3: Calculate moles of glucose

    Moles glucose = mass glucose / molar mass glucose

    Moles glucose = 18.0 grams / 180.156 g/mol

    Moles glucose = 0.0999 moles

    Step 4: Calculate moles of O2

    At STP, 1 mol of gas = 22.4 L. At 37 °C ( = 310 K) and 1 atm, 1 mol = (22.4 L) (310 K/273 K) = 25.4 L.

    Therefore the amount of O₂ here is (7.50 L) / (25.4 L/mol) = 0.295 moles

    Step 5: Determine the limiting reactant

    O2 is the limiting reactant. It will completely be consumed (0.295 moles).

    Glucose is in excess. There will react 0.295/6 = 0.0492 moles

    There will remain 0.0999 - 0.0492 = 0.0507 moles

    Step 6: Calculate moles of water

    For 1 mol glucose, we need 6 moles of O2 to produce 6 moles of CO2 and 6 moles of H2O

    For 0.295 moles of O2, we'll have 0.295 moles of H2O

    Step 7: Calculate mass of water

    Mass of water = moles water * molar mass water

    Mass of water = 0.295 mol * 18.02 g/mol

    Mass of water = 5.32 grams

    There is 5.32 grams of water produced

    Note that the C₆H₁₂O₆ and the O₂ react in a 6:1 ratio. The molar mass of C₆H₁₂O₆ is 180.16 g/mol, so there are (24.5 g) / (180.16 g/mol) = 0.136 mol here. This can react with 6 (0.136) = 0.816 mol O₂.

    Since there are just 0.248 mol O₂, oxygen is the limiting reactant. (1/6) (0.248 mol) = 0.0413 mol glucose is consumed.

    The water is produced in a 6:1 ratio with the glucose consumed, so 0.248 mol are produced.

    (0.248 mol) (18.02 g/mol) = 4.47 g H₂O.
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