HOw much heat is transferred when 8 g of Nitrogen(N2) reacts with 15 g of Hydrogen H2 according to the following equation

N2(g)+3H2--->2NH3(g) enthalpy change is 46.2kJ

 

I have no idea where to start this problem

Use the formula [math]\triangle H_{vap} = n \cdot H_{vap}[/math]

 

You have the total enthalpy change and the moles of the chemicals. You should be able to solve it.

If you wanted to check your work:

 

You have 0.2855 mol (8 g/28.02 g) N2 and 7.44 mol (15 g / 2.016 g) H2. The limiting reagent is the N2 0.29 mol N2 will require 0.2855*3 mole of H2 = 0.8565 mol H2 while using 7.44 mol H2 will require 7.44/3 mol N2 = 2.48 mol N2; thus you dont have enough N2 to use all of the hydrogen (H2 is in excess)

 

Then using the equation eVon gave:

 

Enthalpy = (0.2855 mol)*(46.2 kJ / mol ) = 13.19 kJ for the reaction.

 

 

 

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