2. 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 + 5 O2 -> 3 CO2 + 4 H2O
To calculate how much propane is needed to produce 1000.0 grams of water, we need to use stoichiometry.
From the balanced equation, we can see that the ratio between propane (C3H8) and water (H2O) is 1:4. This means that for every mole of propane burned, 4 moles of water are produced.
The molar mass of water (H2O) is 18.015 g/mol.
To find the moles of water produced from 1000.0 grams, we divide the mass of water by the molar mass:
1000.0 g / 18.015 g/mol = 55.49 mol of water
Since the ratio is 1:4, the number of moles of propane needed is 1/4 of the number of moles of water:
55.49 mol water x (1/4) = 13.87 mol of propane
Therefore, you would need approximately 13.87 moles of propane to produce 1000.0 grams of water.