Calculate the heat of reaction for the incomplete combustion of 67.0g propane producing carbon monoxide gas and water vapour.

how would I go about and do this question?

What are you studying? Do you know how to use the thermodynamic tables (delta Ho, etc).
delta Hrxn = delta H(products) - delta H(reactants)

yes I do!

Look in the tables and find delta H for the products, subtract them from delta H for the the raactants and you will have delta H for the reaction.
You need to write a balanced equation first.
C3H8(g) + O2(g) ==> CO(g) + H2O(g)

Delta H(reaction) will be in kJ/mol and you will compare that with 67 g propane (converted to mols, of course).

To calculate the heat of reaction for the incomplete combustion of propane, we first need to write a balanced equation for the reaction:

C3H8(g) + 5O2(g) → 3CO(g) + 4H2O(g)

Now, let's find the delta H values for the reactants and the products from the thermodynamic tables:

Reactants:
Delta H for C3H8(g) = -104.7 kJ/mol (given)
Delta H for O2(g) = 0 kJ/mol (standard state)

Products:
Delta H for CO(g) = -110.5 kJ/mol (given)
Delta H for H2O(g) = -241.8 kJ/mol (given)

Next, we can calculate the delta H for the reaction using the equation:

Delta H(reaction) = Delta H(products) - Delta H(reactants)

Delta H(reaction) = (3 * Delta H for CO) + (4 * Delta H for H2O) - (Delta H for C3H8 + 5 * Delta H for O2)

Delta H(reaction) = (3 * -110.5 kJ/mol) + (4 * -241.8 kJ/mol) - (-104.7 kJ/mol + 5 * 0 kJ/mol)

Delta H(reaction) = -331.5 kJ/mol + (-967.2 kJ/mol) - (-104.7 kJ/mol)

Delta H(reaction) = -331.5 kJ/mol - 967.2 kJ/mol + 104.7 kJ/mol

Delta H(reaction) = -1194 kJ/mol

Now, to find the heat of reaction for the given mass of propane (67.0 g), we need to convert it to moles. The molar mass of propane (C3H8) is 44.096 g/mol.

Moles of propane = mass of propane / molar mass of propane
Moles of propane = 67.0 g / 44.096 g/mol
Moles of propane ≈ 1.52 mol

Finally, we can calculate the heat of reaction for the given mass of propane:

Heat of reaction = Delta H(reaction) * moles of propane
Heat of reaction = -1194 kJ/mol * 1.52 mol
Heat of reaction ≈ -1814.88 kJ

Therefore, the heat of reaction for the incomplete combustion of 67.0g propane is approximately -1814.88 kJ.