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 produce 1000.0 grams of water, we need to use stoichiometry. From the balanced chemical equation, we see that for every 4 moles of propane used, 4 moles of water are produced. One mole of propane has a mass of 44.11 g, and one mole of water has a mass of 18.02 g.
Therefore, we can set up the following proportion:
4 moles propane / 4 moles water = x grams propane / 1000.0 grams water
Solving for x, we get:
x = (4 moles propane / 4 moles water) * (18.02 g water / 1 mole water) * (1000.0 g water / 1 x)
x = 18040 g propane
Therefore, we would need 18040 grams (or 18.04 kg) of propane to produce 1000.0 grams of water.