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?

1 Answer
Nov 28, 2015

Answer:

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

Explanation:

#"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"#