How many grams of NH3 can be produces from 16.0 of N2 and excess H2?

Answer 1

Question:

How many grams of NH₃ can be produced from 16.0 g of N₂ and excess H₂?

Solution:

First let's balance the reaction

N₂ + H₂ ==> NH₃

N₂ + 3*H₂ ==> 2*NH₃

So for every N₂ we get 2*NH₃

Given:

N₂ = 16.0 g

From the Periodic Table:

1 mole N = 14.007 g

1 mole H = 1.008 g

How many moles of N do we have?

16.0 g of N₂ * (1 mole N / 14.007 g) = 1.14 moles of N

This is the same number of moles for NH₃.

NH3 weighs N + 3*H

= 14.007 g + 3*1.008 g = 17.03 g / mole

Multiple to solve:

1.14 moles of NH₃ * 17.03 g / mole = 19.42 g

THEREFORE

19.42 g of NH₃ can be produced