Question

How many liters of water can be boiled by burning 1 kg of propane?

Round your answer to the nearest whole number.

Answer 1

2037 L

Step-by-step explanation:

Step 1:

The balanced equation for the reaction involving the burning of propane. This is given below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Step 2:

Determination of the number of mole in 1kg of propane (C3H8).

Mass of C3H8 = 1kg = 1000g

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

Number of mole = Mass/Molar Mass

Number of mole of C3H8 = 1000/44

Number of mole of C3H8 = 22.73 moles

Step 3:

Determination of the number of mole of water produced by burning 1 kg of propane (C3H8). This is illustrated below:

From the balanced equation above,

1 mole of C3H8 boiled 4 moles of H2O.

Therefore, 22.73 moles of C3H8 will produce = 22.73 x 4 = 90.92 moles of H2O.

Step 4:

Determination of the volume of H2O. This is illustrated below:

1 mole of a gas occupy 22.4L.

Therefore, 90.92 moles of H2O will occupy = 90.92 x 22.4 = 2037 L.

Therefore, 2037 L of water is boiled by burning 1kg of propane (C3H8)