Respuesta :
Answer:
a) O₂ is limiting reactant
b) 0.0645 moles of CO₂
c) 1.55g of H₂O
d) 13.85g C₃H₈
Explanation:
The balanced reaction is:
a C₃H₈ + 5 O₂ → 3CO₂ + 4H₂O
1 mole of C₃H₈ reacts per 5 moles of O₂
14.8g of C₃H₈ and 3.44g of O₂ are:
14.8g × (1mol / 44.1g) = 0.3356moles C₃H₈
3.44g × (1mol / 32g) = 0.1075moles O₂
A complete reaction of 0.3356moles of C₃H₈ requires:
0.3356moles of C₃H₈ × (5mol O₂ / 1mol C₃H₈) = 1.678 mol O₂
As you start with just 0.1075 moles, O₂ is limiting reactant
b 0.1075moles O₂ produce:
0.1075moles O₂ × (3mol CO₂ / 5 mol O₂) = 0.0645 moles of CO₂
c Mass of H₂O produced from 0.1075moles O₂ is:
0.1075moles O₂ × (4mol H₂O / 5 mol O₂) × (18.01g / 1 mol H₂O) =
1.55g of H₂O
d 0.1075 moles O₂ react completely with:
0.1075 moles O₂ × (1mol C₃H₈ / 5mol O₂) = 0.0215mol C₃H₈.
As you have 0.3356 moles C₃H₈, moles that don't react are:
0.3356moles C₃H₈ - 0.0215mol C₃H₈ = 0.3141 moles C₃H₈. In grams:
0.3141 moles C₃H₈ (44.1g / 1mol) = 13.85g C₃H₈
Answer:
a) Oxygen is the limiting reactant.
b) 0.0645 moles CO2= 2.84 grams CO2
c) Mass H2O = 1.55 grams
d) Mass propane remaining = 13.85 grams
Explanation:
Step 1: Data given
MAss of propane (C3H8) = 14.8 grams
Mass of oxygen (O2) = 3.44 g/mol
Molar mass of O2 = 32.0 g/mol
Molar mass of propane = 44.1 g/mol
Step 2: The balanced equation
C3H8 + 5O2 ---->3CO2 + 4H2O
Step 3: Calculate moles
Moles =mass / molar mass
Moles propane = 14.8 grams / 44.1 g/mol
Moles propane = 0.3356 moles
Moles oxygen = 3.44 grams / 32.0 g/mol
Moles oxygen = 0.1075 moles
Step 4: Calculate the limiting reactant
For 1 mol C3H8 we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O
Oxygen is the limiting reactant. It will completely be consumed ( 0.1075 moles). Propane is in exess. There will react 0.1075/5 = 0.0215 moles
There will remain 0.3356 - 0.0215 = 0.3141 moles
step 5: Calculate mass of propane remaining
Mass propane = moles * molar mass
Mass propane remaining = 0.3141 moles * 44.1 g/mol
Mass propane remaining = 13.85 grams
Step 6: Calculate moles products
For 1 mol C3H8 we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O
For 0.1075 moles O2 we'll have 3/5 * 0.1075 = 0.0645 moles CO2
For 0.1075 moles O2 we'll have 4/5 * 0.1075 = 0.086 moles H2O
Step 7: calculate mass products
Mass = moles * molar mass
Mass CO2 = 0.0645 moles * 44.01 g/mol
Mass CO2 = 2.84 grams
Mass H2O = 0.086 grams * 18.02 g/mol
Mass H2O = 1.55 grams
