How many grams of hydrogen will react with excess nitrogen to form 10.0 moles ammonia? How may moles of hydrogen must react with 56 grams of nitrogen?    

Expert Answers

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Hydrogen and nitrogen react to form ammonia. The relevant chemical equation can be written as:

`N_2 + 3H_2 -> 2NH_3`

Here 1 mole of nitrogen reacts with 3 moles of hydrogen to form 2 moles of ammonia. 

In the first case, excess nitrogen is present. Using stoichiometry,

2 moles ammonia requires 3 moles hydrogen,

1 mole of ammonia will require 3/2 moles of hydrogen

and 10 moles of ammonia will require 3/2 x 10 = 15 moles of hydrogen.

The molar mass of hydrogen is 2 g.

Thus, 15 moles x 2 g/mole = 30 g of hydrogen is required to make 10 moles of ammonia.

Molar mass of nitrogen is 28 g and thus, 56 g of nitrogen contains 56 g / 28 (g/mole) =  2 moles.

Using the well balanced chemical equation for formation of ammonia, 1 mole of nitrogen reacts with 3 moles of hydrogen and thus, 2 moles of nitrogen will react with 2 x 3 = 6 moles of hydrogen.

Hope this helps. 


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