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?
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.