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How many moles of nh3 can be produced from 13.5 mol of h2 and excess n2?

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Topic starter Posted : 06/02/2021 2:07 pm
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This question is based on “Limiting reagent” concept. It explains that the reactant taht is present in lesser quantity or concentration than required for completing the reaction along with other reactants is known as the limiting reagent. 

This reagent limits the production of final products formed and also decides the quantity of product formed in the reaction. Such limit reagents are entirely consumed in a chemical reaction. 

Here Hydrogen is the limit reagent. To find out the number of moles formed, we will compare the mole ratio of the reactants we have with the ones used in the reaction. 

N2 + 3H2→ 2NH3

This is a balanced chemical equation for Ammonia which shows that 3 moles of Hydrogen produce 2 moles of Ammonia. 

So to calculate the moles of Nh3 formed when we have 13.5 moles of Hydrogen.

We will use this method,

nNH3 = 13.5 moles H2 * 2 moles NH3/ 3 moles H2

Now place all the values in the equation to find out the answer.

= 13.5 x 2 / 3

= 9 moles

Thus, 13.5 moles of Hydrogen will produce 9 moles of Ammonia in the presence of excess Nitrogen. 

Posted : 06/02/2021 2:08 pm