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?

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

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa