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QUESTION

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

"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"