Consider:

C6H12)6 + 6O2-> 6CO2 + 6H2O + Energy

a)If I have 90 grams glucose and all is consumed by oxygen, then how many grams of water will be produced?

b)How many grams of oxygen will be consumed?

c)How many grams of sugar will be burned off if 85 grams of O2 is consumed?

Here is a worked example problem. Just follow the steps.

http://www.jiskha.com/science/chemistry/stoichiometry.html

That helps me with part a but are the other 2 parts done the same way?

Yes. All are simple (that means not limiting reagent) stoichiometry problems.All of them are done the same way.

To solve these questions, we need to use the balanced chemical equation provided, which shows the stoichiometric relationship between the reactants and products.

a) The balanced equation states that for every 1 mole of glucose (C6H12O6) consumed, 6 moles of water (H2O) are produced. To determine the grams of water produced, we need to convert the given mass of glucose to moles and then use the stoichiometry to find the mass of water produced.

1 mole of glucose (C6H12O6) has a molar mass of approximately 180.16 g/mol. Therefore, 90 grams of glucose is equal to:

90 g / 180.16 g/mol = 0.4988 moles

According to the balanced equation, 1 mole of glucose produces 6 moles of water. Therefore, 0.4988 moles of glucose would produce:

0.4988 moles * 6 moles H2O = 2.9928 moles of H2O

To convert this back to grams, we multiply by the molar mass of water, approximately 18.02 g/mol:

2.9928 * 18.02 g/mol = 53.91 grams of water

Therefore, approximately 53.91 grams of water will be produced.

b) The balanced equation shows that for every 1 mole of glucose (C6H12O6) consumed, 6 moles of oxygen (O2) are required. To determine the grams of oxygen consumed, we need to convert the given mass of glucose to moles and then use the stoichiometry to find the mass of oxygen consumed.

Using the same calculation as in part a, 90 grams of glucose is equal to 0.4988 moles of glucose.

According to the balanced equation, 1 mole of glucose requires 6 moles of O2. Therefore, 0.4988 moles of glucose would require:

0.4988 moles * 6 moles O2 = 2.9928 moles of O2

To convert this back to grams, we multiply by the molar mass of oxygen, approximately 32.00 g/mol:

2.9928 * 32.00 g/mol = 95.77 grams of oxygen

Therefore, approximately 95.77 grams of oxygen will be consumed.

c) To determine how many grams of glucose will be burned off if 85 grams of O2 is consumed, we need to use the stoichiometry of the balanced equation.

First, we convert the given mass of oxygen to moles using its molar mass of approximately 32.00 g/mol:

85 g / 32.00 g/mol = 2.6563 moles of O2

According to the balanced equation, 1 mole of glucose requires 6 moles of O2. Therefore, 2.6563 moles of O2 would require:

2.6563 moles * (1 mole C6H12O6 / 6 moles O2) ≈ 0.4427 moles of C6H12O6

To convert this back to grams, we multiply by the molar mass of glucose, approximately 180.16 g/mol:

0.4427 * 180.16 g/mol ≈ 79.71 grams of glucose

Therefore, approximately 79.71 grams of sugar will be burned off if 85 grams of O2 is consumed.