Question

Propane (C3H8) burns in oxygen to produce carbon dioxide gas and water vapor. (a) Write a balanced equation for this reaction. Include all reaction states. → (b) Calculate the number of liters of carbon dioxide measured at STP that could be produced from 7.65 g of propane. L

Answer 1

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

Answer 2

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