How many grams carbon dioxide is made when 42 grams of propane is burned with 115 grams of oxygen how many grams of water is made?

Answer 1

First you need to write a balanced equation. You are given that propane undergoes a combustion reaction that produces carbon dioxide and water.

Unbalanced: C3H8 + O2 ---> CO2 + H2O

Balanced: C3H8 + 5O2 ---> 3CO2 + 4H2O

Givens:

42.0 grams C3H8 (Molecular mass 44.0 g)

115.0 grams O2 (Molecular mass 32.0 g)

Molecular mass of CO2: 44.0 g

Mole ratio 1:5:3:4 (C3H8:O2:CO2:H2O)

Then you need to find which of the reactants are the limiting reactant (lowest value) and which is the excess reactant. The limiting reactant is what you will base the rest of the problem on. To do this, you convert each measurement to moles from grams.42.0 g C3H8 / (44.0 g) = .955 moles C3H8 115.0 g

Answer 2

First you need to write a balanced equation. You are given that propane undergoes a combustion reaction that produces carbon dioxide and water.

Unbalanced: C3H8 + O2 ---> CO2 + H2O

Balanced: C3H8 + 5O2 ---> 3CO2 + 4H2O

Givens:

42.0 grams C3H8 (Molecular mass 44.0 g)

115.0 grams O2 (Molecular mass 32.0 g)

Molecular mass of CO2: 44.0 g

Mole ratio 1:5:3:4 (C3H8:O2:CO2:H2O)

Then you need to find which of the reactants are the limiting reactant (lowest value) and which is the excess reactant. The limiting reactant is what you will base the rest of the problem on. To do this, you convert each measurement to moles from grams.42.0 g C3H8 / (44.0 g) = .955 moles C3H8 115.0 g× (5 moles) / (32.0 g) = 18.0 moles O2 Because propane is the limiting reactant (less of it than oxygen), the amount of water produced can only be as much as the propane. Take the amount of propane and use stoichiometry to find the amount of water produced in grams.

42.0 g C3H8 / (44.0 g) × (4 moles) × (18 g) = 68.7 grams H2O