Answer:
11.6 L will be the number of liters of carbon dioxide measured at STP.
Explanation:
The balanced equation for this combustion reaction is:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
where 1 mol of propane reacts to 5 moles of oxygen in order to produce 3 moles of carbon dioxide and 4 moles of water.
We assume the oxygen in excess, so the limiting reagent is the propane. Now, we determine the moles: 7.65 g . 1 mol/ 44 g = 0.174 moles
Ratio is 1:3. 1 mol of propane can produce 3 moles of CO₂
Therefore, 0.174 moles will produce (0.174 . 3) / 1 = 0.521 moles of CO₂
As 1 mol of gas is contained in 22.4L at STP conditions, we propose
22.4L / 1 mol = V₂ / 0.521 mol
22.4 L / 1 mol . 0.521 mol = V₂ → 11.6 L
Answer:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O (g)
11.6 L of CO2 will be produced
Explanation:
Step 1: Data given
Mass of propane = 7.65 grams
Molar mass propane = 44.1 g/mol
Burning = combustion reation = adding O2. The products will be CO2 and H2O
Step 2: The balanced equation
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O (g)
Step 3: Calculate moles propane
Moles propane = 7.65 grams / 44.1 g/mol
Moles propane = 0.173 moles
Step 4: Calculate moles CO2
For 1 mol propane we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O
For 0.173 moles propane we'll have 3*0.173 = 0.519 moles CO2
Step 5: Calculate volume of CO2
1 mol = 22.4 L
0.519 moles = 22.4 L * 0.519 = 11.6 L
11.6 L of CO2 will be produced
