Answer:
3.60 grams of Hydrogen
Step-by-step explanation:
This is a problem of stoichiometry so we first need the balance chemical reaction and then convert the quantities given to mol and determine our limiting reagent to calcuale the amount of the excess reagent if any.
N₂ + 3 H₂ ------------------------------ 2 NH₃
MW N₂ = 28.0 g/mol mol N₂ = 30g x 1mol/28.0 g = 1.07 mol
MW H₂ = 2.01 g/mol molH₂= 10 g x 1 mol /2.01 = 4.98 mol
mol H₂ required to completely react with the N₂ :
3mol H₂/ 1 mol N₂ x 1.07 mol N₂ = 3.21 mol H2
we have plenty of H₂.
mol N₂ required to completely react with the H₂ :
1 mol N₂/ 3 mol H₂ x 4.98 mol H₂ = 1.66 mol N₂
but we have only 1.07 mol N₂, therefore our limiting reagent is nitrogen and hydrogen is in excess. The amount in excess is:
5 mol H₂ - 3.21 mol H₂ reacted = 1.79 mol H₂
1.79 mol H₂ times 2 g/mol will give us the amount of hydrogen left unconsumed:
1.79 mol x 2.01 g/mol = 3.60g H₂