Question

Propane is often used to heat homes. The combustion of propane follows the following reaction: C3H8(g) + 5O2(g)  3CO2 (g) + 4H2O(g), ΔH° = –2044 kJ. How many grams of propane must be reacted by this reaction to release 7563 kJ of heat?

Answer 1

Answer:

To release 7563 kJ of heat, we need to burn 163.17 grams of propane

Explanation:

Step 1: Data given

C3H8 + 5O2 -----------> 3CO2 + 4H2O      ΔH° = –2044 kJ

This means every mole C3H8

Every mole of C3H8 produces 2044 kJ of heat when it burns (ΔH° is negative because it's an exothermic reaction)

Step 2: Calculate the number of moles to produce 7563 kJ of heat

1 mol = 2044 kJ

x mol = 7563 kJ

x = 7563/2044 =  3.70 moles

To produce 7563 kJ of heat we have to burn 3.70 moles of C3H8

Step 3: Calculate mass of propane

Mass propane = moles * Molar mass

Mass propane = 3.70 moles * 44.1 g/mol

Mass propane = 163.17 grams

To release 7563 kJ of heat, we need to burn 163.17 grams of propane

Answer 2

The mass of propane needed by the reaction to release 7563 KJ of heat energy is 162.8 g

Balanced equation

C₃H₈ + 5O₂ —> 3CO₂ + 4H₂O ΔH° = –2044 KJ

Molar mass of propane C₃H₈ = (3×12) + (1×8) = 44 g/mol

Mass of propane C₃H₈ from the balanced equation = 1 × 44 = 44 g

Heat released = 2044 KJ

SUMMARY

From the balanced equation above,

2044 KJ of heat was released by the reaction of 44 g of propane

How to determine the mass of propane needed

From the balanced equation above,

2044 KJ of heat was released by the reaction of 44 g of propane

Therefore,

7563 KJ of heat energy will be release by = (7563 × 44) / 2044 = 162.8 g of propane

Thus, 162.8 g of propane is needed for the reaction

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