##”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
- Divide ##”8.064 g”## of glucose by its molar mass to get moles glucose.
- 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.
- 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”##