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)
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Ответ:
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
Ответ:
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