Question

For the following reaction, 10.0 grams of nitrogen gas are allowed to react with 1.19 grams of hydrogen gas . nitrogen(g) + hydrogen(g) ammonia(g) What is the maximum mass of ammonia that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams

Answer 1

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

Step-by-step 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