Question

Be sure to answer all parts. Ammonia is produced by the reaction of nitrogen and hydrogen according to the equation N2(g) + 3H2(g) + 2NH3(g) Calculate the mass of ammonia produced when 33.0 g of nitrogen react with 11.2 g of hydrogen. Which is the excess reactant and how much of it will be left over when the reaction is complete? o hydrogen nitrogen

Answer 1

Formation of ammonia

N2 (g) + 3H2 (g) = 2NH3 (g)

Given that nitrogen = 33.0 g

Hydrogen = 11.2 g

Number of moles, N2= amount in g / molar mass

= 33.0 g/ 28.0134 g/mol

=1.18 moles N2

Moles of H2= 11.2 g /2.016 g/ mole

= 5.56 moles

Moles of H2= 1.18 moles N2 * 3/1

= 3.54 moles H2

Moles of N2= 5.56 moles*1/3

=1.85 mole N2

Thus nitrogen is limiting agent

The limiting agent has following properties:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mole.

1.18 moles N2 *2 mole NH3/ 1 mole N2

= 2.36 mole NH3

Amount in g = number of mole s* molar mass

= 2.36 mole NH3*17.031 g/mol

= 40.19 g

Hydrogen is present in excess

Moles of excess= total moles – used moles

= 5.56 moles – 3.54 moles

= 2.02 moles

Amount in g = number of moles * molar mass

= 2.02 moles *2.016 g/ moles

= 4.07 g excess hydrogen