What volume of propane (C3H8) is required to produce 165 liters of water according to the following reaction? (All gases are at the same temperature and pressure.) propane (C3H8) (g) + oxygen (g) carbon dioxide (g) + water (g)

We need 41.2 L of propane

Explanation:

Step 1: Data given

volume of H2O = 165 L

Step 2: The balanced equation

C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)

Step 3: Calculate moles of H2O

1 mol = 22.4 L

165 L = 7.37 moles

Step 4: Calculate moles of propane

For 1 mol C3H8 we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O

For 7.37 moles H2O we need 7.37/4 = 1.84 moles propane

Step 5: Calculate volume of propane

1 mol = 22.4 L

1.84 moles = 41.2 L

We need 41.2 L of propane