Answer:
a) H₂ is less than the required amount, hence known to be as the limiting reactant in this reaction.
b) 6.75 g of NH₃ excess reagent remains after the reaction is complete
Explanation:
N₂ + 3 H₂ ------------- 2 NH₃
Given amount of N₂ = 10 g
Given amount of H₂ = 1.19 g
Moles of N₂ = mass/ molar mass
= 10g/ 28g/mol = 0.357 moles
Moles of H₂ = mass/ molar mass
= 1.19g/ 2g/mol = 0.595 moles
We know that
1 mole of N₂ Reacts = 3 moles of H₂
0.357 moles of H₂ = x moles of H₂
X = 0.357 moles N₂ x 3 moles H₂/ 1 mole N₂
= 1.071 moles of H₂
H₂ is less than the required amount, hence known to be as the limiting reactant in this reaction.
B) For mass of the excess reagent remains, first find the number of moles
3 mole of H₂ produces 2 moles of NH₃
0.595 moles of H₂ produces x mol NH3
X= 0.595 mol H₂ x 2 mol NH₃ /3 mol H₂
= 0.396 mole NH₃
Mass = moles x molar mass
= 0.396 x 17.031
= 6.75 g NH₃
6.75 g of NH₃ excess reagent remains after the reaction is complete
