32 grams of propane (c3h8) is burned in excess oxygen gas to produce how many grams of water?
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Ответ:
There will be produced 52.29 grams of water
Explanation:
Step 1: Data given
Mass of propane = 32.00 grams
Molar mass of propane = 44.1 g/mol
Oxygen = in excess
Step 2: The balanced equation
C3H8 + 5O2 → 3CO2 + 4H2O
Step 3: Calculate moles of propane
Moles propane = Mass propane / molar mass propane
Moles propane = 32.00 grams / 44.10 g/mol
Moles propane = 0.7256 moles
Step 4: Calculate moles of water
The limiting reactant is propane.
For 1 mol of propane consumed, we need 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O
For 0.7256 moles of propane consumed we'll have 4* 0.7256 = 2.902
Step 5: Calculate mass of H2O
Mass H2O = moles H2O * molar mass H2O
Mass H2O = 2.902 moles * 18.02 g/mol
Mass H2O = 52.29 grams
There will be produced 52.29 grams of water
Ответ:
D
Step-by-step explanation: