The Equation For The Combustion Of Glucose Is: C6H12O6(s) + 6O2(g ...

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? Chemistry Chemical Reactions Chemical Reactions and Equations

1 Answer

Meave60 Nov 28, 2015

#"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"#

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