The equation for respiration reaction is given by:

C6H12O6 + 6O2 ------> 6CO2 + 6H2O + Energy

Calculate the amount of water formed from oxidizing 18.0 grams of glucose in the presence of excess oxygen.

How many moles of glucose is 18grams? The balanced equation is that it will yield six times that number of moles of water.

To calculate the amount of water formed from oxidizing 18.0 grams of glucose, we need to use the stoichiometry of the balanced equation for respiration.

According to the equation:
1 mole of glucose (C6H12O6) reacts with 6 moles of oxygen (O2) to produce 6 moles of carbon dioxide (CO2), 6 moles of water (H2O), and release energy.

First, we need to convert the given mass of glucose into moles.

The molar mass of glucose (C6H12O6) can be calculated by adding up the atomic masses of the elements in its formula:
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol

Molar mass of glucose (C6H12O6) = (6 * C) + (12 * H) + (6 * O) = (6 * 12.01 g/mol) + (12 * 1.01 g/mol) + (6 * 16.00 g/mol) = 180.18 g/mol

Next, we can use the molar ratio of water (H2O) to glucose (C6H12O6) from the balanced equation, which is 6:1.

Moles of glucose = (mass of glucose) / (molar mass of glucose)
Moles of glucose = 18.0 g / 180.18 g/mol = 0.0999 mol (approximately)

Since the molar ratio of water (H2O) to glucose (C6H12O6) is 6:1, the amount of water formed will be 6 times the number of moles of glucose.

Moles of water = (moles of glucose) * 6
Moles of water = 0.0999 mol * 6 = 0.5994 mol (approximately)

Finally, we can convert moles of water into grams by multiplying by the molar mass of water (H2O).

Mass of water = (moles of water) * (molar mass of water)
Mass of water = 0.5994 mol * 18.02 g/mol = 10.8 g (approximately)

Therefore, the amount of water formed from oxidizing 18.0 grams of glucose in the presence of excess oxygen is approximately 10.8 grams.