Ans : Number of moles of nitrogen = 1000.0 / 28.0134 = 35.697 moles Number of mo

Question

Ans :

Number of moles of nitrogen = 1000.0 / 28.0134 = 35.697 moles

Number of moles of hydrogen = 500.0 / 2.01588 = 248 moles

Nitrogen is the limiting reagent here.

Number of moles of ammonia produced will be : 2 x 35.697 = 71.39 moles

Mass of ammonia = 71.39 x 17.03052

= 1215.88 grams

Ans b:

Here the excess reactant is hydrogen.

Number of moles of hydrogen utilised will be = 35.697 x 3 = 107.09 moles

Amount of hydrogen utilised = 107.09 x 2.01588

= 215.88 grams

So mass of hydrogen left = 500.0 - 215.88

= 284.12 grams

Explanation / Answer

Ammonia is produced commercially using the Haber Process, according to the following reaction: N2 + 3 H22 NH3 a) What is the maximum mass of ammonia that can be produced from a mixture of 1000.0 grams 20 of nitrogen and 500.0 grams of hydrogen? What mass of the excess reactant remains after the reaction is completed? b) Cancel

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