Question

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?

Answer 1

`"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"`