N2+3H2-2NH3
calculate the number of grams of NH3 are formed when 3.0 mole of N2 is reacted with 6.7 moles of H2

102 grams of ammonia is formed when 3 moles of nitrogen and 6.7 moles of hydrogen reacts.

Explanation:

The equation given is of Haeber's process in which the nitrogen is limiting factor in the ammonia formation and hydrogen if in excess gets delimited.

We know that 1 mole of Nitrogen gives 2 moles of ammonia.

We have 3 moles of nitrogen here,

So, 6 moles of ammonia will be form

so from the formula

no of moles=mass/atomic mass

           mass= no. of moles*atomic mass

                     = 6*17

                       = 102 grams of ammonia will be formed.

So, 6 moles or 102 grams of ammonia is formed when 3 mole of nitrogen and 6.7 mole of hydrogen reacts.

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

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