32 grams of propane (c3h8) is burned in excess oxygen gas to produce how many grams of water?

There will be produced 52.29 grams of water

Explanation:

Step 1: Data given

Mass of propane = 32.00 grams

Molar mass of propane = 44.1 g/mol

Oxygen = in excess

Step 2: The balanced equation

C3H8 + 5O2 → 3CO2 + 4H2O

Step 3: Calculate moles of propane

Moles propane = Mass propane / molar mass propane

Moles propane = 32.00 grams / 44.10 g/mol

Moles propane = 0.7256 moles

Step 4: Calculate moles of water

The limiting reactant is propane.

For 1 mol of propane consumed, we need 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O

For 0.7256 moles of propane consumed we'll have 4* 0.7256 = 2.902

Step 5: Calculate mass of H2O

Mass H2O = moles H2O * molar mass H2O

Mass H2O = 2.902 moles * 18.02 g/mol

Mass H2O = 52.29 grams

There will be produced 52.29 grams of water

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation: