Answer:
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)