The equation for the combustion of glucose is: C6H12O6(s) + 6O2(g) –>6CO2(g) + 6H2O(g). How many grams of H2O will be produced when 8.064g of glucose is burned?

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##”4.838 g H”_2″O”## will be formed from the combustion of ##”8.064 g C”_6″H”_12″O”_6″##.

##”C”_6″H”_12″O”_6″ + 6O”_2″####rarr####”6CO”_2″ + 6H”_2″O”##

Molar Masses of Glucose and Water ##”C”_6″H”_12″O”_6″:####”180.15588 g/mol”## ##”H”_2″O”:####”18.01528 g/mol”##https://pubchem.ncbi.nlm.nih.gov

  1. Divide ##”8.064 g”## of glucose by its molar mass to get moles glucose.
  2. Multiply times the mol ratio ##(6″mol H”_2″O”)/(1″mol 8C”_6″H”_12″O”_6″)## from the balanced equation to get moles water.
  3. Multiply times the molar mass of ##”H”_2″O”## to get mass of water.

##8.064cancel(“g C”_6″H”_12″O”_6)xx(1cancel(“mol C”_6″H”_12″O”_6))/(180.15588cancel(“g C”_6″H”_12″O”_6))xx(6cancel(“mol H”_2″O”))/(1cancel(“mol C”_6″H”_12″O”_6))xx(18.01528″g H”_2″O”)/(1cancel(“mol H”_2″O”))=”4.838 g H”_2″O”##