When burning 180g of glucose in the presence of 192g of oxygen, water and carbon dioxide are produced?

The burning of glucose in presence of oxygen is called as combustion

The balanced reaction is

C6H12O6 + 6O2  ---> 6CO2 + 6H2O

Thus for each mole of glucose we need six moles of oxygen molecules

molar mass of glucose is 180g / mole

The moles of glucose present = mass / molar mass = 180 / 180 = 1 mole

The moles of oxygen present = mass / molar mass = 192 / 32 = 6 moles

Thus the reaction will go to completion and one mole of glucose will react with six moles of O2 to give six moles of CO2 and six moles of H2O

Mass of CO2 produced = moles X molar mass = 6 x 44 = 264 g

Mass of H2O produced = moles X molar mass = 6 x 18 = 108 g

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

0.88m hope i am correct : )