Question

the amount of water produced by combustion of 286 gram of propane is

Answer 1

Given:

Propane(C₃H₈) undergoes combustion

Mass of propane = 286 g

To Find:

The amount of water produced by the combustion of 286 g of propane

Solution:

• The combustion of propane occurs according to the equation:

                   C₃H₈ + 5O₂ →  3CO₂ + 4H₂O

• From this balanced equation, it is clear that

(1 mole of propane) + (5 moles of oxygen) → (3 moles of carbon dioxide) +  (4 moles of water).

• To solve the question, we will follow these steps:

Step 1. Find the number of moles of propane.

• Molar mass of C₃H₈ = 3(12) + 8(1)

                                          = 44 g

• Given mass of propane = 286 g

∴ Number of moles of propane = $\frac{286}{44}$ = 6.5

Step 2. Calculate the number of moles of water produced.

• From the equation of combustion of propane, we know

                  1 mole of propane gives 4 moles of water

Thus, 6.5 moles of propane will give⇒

                         Moles of water = 6.5 × 4

                                                   = 26

Thus, 26 moles of water are produced.

Step 3. Calculate the mass of water.

• The molar mass of water = 18 g
• Number of moles of water calculated = 26

∴ Mass of 26 moles of water = 18 × 26

                                                = 468 g

Thus, the amount of water produced by the combustion of 286 g of propane is 468 g.