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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?

Respuesta :

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

Eduard22sly

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

Learn more enthalpy change:

https://brainly.com/question/11967318

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