Propane(C3H8), the major component of LP game, is often used as heating and cooking fuels in homes. When propane is burned( reacted with oxygen), CO2(g) and H2O(g) are produced.

1) Write an equation for the reaction and calculate the enthalpy change for the combustion of propane.
2) How much heat can be produced by the combustion of 50.0 g propane?

C3H8 + 5O2 ==> 3CO2 + 4H2O

I suppose you want the standard enthalpy change using delta Hf values.
delta Hrxn = (n*DHfproducts)-(n*DHfreactants)

whatever is easier!

Look up delta Hf for CO2 and multiply by n. Look up delta Hf for H2O an multiply by n. Look up delta Hf for C3H8 and multiply by n. Then

DHrxn = (n*DHproducts)-(n*DHreactants). That gives you delta H for the combustion of propane.
For the second part,
q = DHrxn x (50g/molar mass propane)

To answer these questions, we need to follow a few steps:

1) Write the balanced chemical equation for the combustion of propane:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Here, one molecule of propane reacts with five molecules of oxygen to produce three molecules of carbon dioxide and four molecules of water.

2) Calculate the enthalpy change (ΔH) for the combustion of propane:
To calculate the enthalpy change, we need to determine the standard enthalpy of formation (ΔHf) for each species involved. The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states, where standard states refers to temperature and pressure conditions.

Using the table of standard enthalpies of formation, we find:
ΔHf(C₃H₈) = -104.7 kJ/mol
ΔHf(CO₂) = -393.5 kJ/mol
ΔHf(H₂O) = -241.8 kJ/mol
ΔHf(O₂) = 0 kJ/mol

Next, we can calculate the enthalpy change for the reaction:
ΔH = (ΔHf(products) - ΔHf(reactants))
= [3 * ΔHf(CO₂) + 4 * ΔHf(H₂O)] - [ΔHf(C₃H₈) + 5 * ΔHf(O₂)]
= [3 * (-393.5 kJ/mol) + 4 * (-241.8 kJ/mol)] - [(-104.7 kJ/mol) + 5 * 0 kJ/mol]
= -2220.3 kJ/mol

Therefore, the enthalpy change for the combustion of propane is -2220.3 kJ/mol.

3) Calculate the amount of heat produced by the combustion of 50.0 g of propane:
To do this, we need to use the molar mass of propane and convert grams to moles:
Molar mass of propane (C₃H₈) = (3 * Atomic mass of carbon) + (8 * Atomic mass of hydrogen)
= (3 * 12.01 g/mol) + (8 * 1.008 g/mol)
= 44.1 g/mol

Number of moles of propane = Mass of propane / Molar mass of propane
= 50.0 g / 44.1 g/mol
= 1.13 mol

Now, we can use the enthalpy change calculated earlier to determine the heat produced by the combustion of propane:
Heat = ΔH * Number of moles of propane
= -2220.3 kJ/mol * 1.13 mol
= -2508.4 kJ

Therefore, the combustion of 50.0 g of propane can produce approximately 2508.4 kJ of heat.