Propane + oxygen yields carbon dioxide and water
Label what type of reaction (synthesis, decomposition, single replacement, double replacement or combustion)
Write the balanced chemical equation
How much propane would you need to produce 1000.0 grams of water?
Type of reaction: Combustion
Balanced chemical equation: C3H8 + 5O2 -> 3CO2 + 4H2O
To calculate how much propane is needed to produce 1000.0 grams of water, we need to determine the mole ratio between propane and water in the balanced equation. From the equation, we know that for every 4 moles of H2O produced, we need 1 mole of propane.
Molar mass of water (H2O): 18.015 g/mol
Molar mass of propane (C3H8): 44.097 g/mol
Using the mole ratio, we can set up a proportion:
1 mole propane / 4 moles water = x moles propane / 1000.0 g water
Solving for x:
x = (1 mole propane / 4 moles water) * (1000.0 g water / 18.015 g/mol) * (1 mol C3H8 / 3 mol H2O) * (44.097 g/mol / 1 mol C3H8)
x = 306.34 g propane
Therefore, 306.34 grams of propane are needed to produce 1000.0 grams of water in this reaction.