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stoichiometry

Screenshot 2021-06-23 010602.png
the answer is A
i know there should be 0.5 mol of reactant but how do you work out moles of product
Original post by taxonomy.
Screenshot 2021-06-23 010602.png
the answer is A
i know there should be 0.5 mol of reactant but how do you work out moles of product

You begin with 1 mole of reactant (ammonia) and 0 moles of product (nitrogen and hydrogen).

Then, 50% of ammonia dissociates. This means 50% of the original ammonia is breaking down to form nitrogen and hydrogen. 50% of the original number of moles (1) is 0.5.

Using the stoichiometery, you know that 1 mole of ammonia gives 0.5 moles nitrogen and 1.5 moles hydrogen. So, by the same ratio, the 0.5 moles of ammonia that reacted would produce 0.25 moles of nitrogen and 0.75 moles of hydrogen.

Now let us look at the amount of substances at equilibrium. With the initial 1 mole of ammonia, 50% of it dissociated, bringing it to 0.5 moles of ammonia at equilibrium. Initially you had 0 moles of nitrogen, but now you have 0.25. And finally you had 0 moles of hydrogen but now you have 0.75. Add all of the moles of gases available at equilibrium (0.5 + 0.25 + 0.75) and you have your answer
Reply 2
Original post by qwert7890
You begin with 1 mole of reactant (ammonia) and 0 moles of product (nitrogen and hydrogen).

Then, 50% of ammonia dissociates. This means 50% of the original ammonia is breaking down to form nitrogen and hydrogen. 50% of the original number of moles (1) is 0.5.

Using the stoichiometery, you know that 1 mole of ammonia gives 0.5 moles nitrogen and 1.5 moles hydrogen. So, by the same ratio, the 0.5 moles of ammonia that reacted would produce 0.25 moles of nitrogen and 0.75 moles of hydrogen.

Now let us look at the amount of substances at equilibrium. With the initial 1 mole of ammonia, 50% of it dissociated, bringing it to 0.5 moles of ammonia at equilibrium. Initially you had 0 moles of nitrogen, but now you have 0.25. And finally you had 0 moles of hydrogen but now you have 0.75. Add all of the moles of gases available at equilibrium (0.5 + 0.25 + 0.75) and you have your answer

thankyou

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