Question

Q1.The table below shows data for the four hydrocarbons ethyne, propyne, propene and propane. AHc is the

Answer 1

Answer:

$\large \boxed{\text{A. Ethyne}}$

Explanation:

It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.

Then you can consider it to be "moles" of "kJ"  

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r:               26

              2H-C≡C-H + … ⟶   1300 kJ + …

m/g:              2.0

1. Moles of HC≡CH

$\text{Moles of HC \equiv CH} = \text{2.0 g} \times \dfrac{\text{1 mol}}{\text{26 g}} = \dfrac{1}{13}\text{ mol}$

2. Heat released

The molar ratio is 1300 kJ:1 mol HC≡CH

$\text{Heat} = \dfrac{1}{13} \text{ mol HC \equiv CH } \times \dfrac{\text{1300 kJ}}{\text{1 mol HC \equiv CH}} = \textbf{100 kJ}\\\\\text{The reaction of 2.0 g of \large \boxed{\textbf{HC \equiv CH}} produces 100 kJ.}$