Question

Consider the reaction $N_2 + 3 H_2 \rightarrow 2 NH_3$. How much NH₃ can be produced from the reaction of 74.2 g of N₂ and 14.0 moles of H₂? a. 1. 1.59 × 10²⁴ molecules b. 2. 1.69 × 10²⁵ molecules c. 3. 3.19 × 10²⁴ molecules d. 4. 1.26 × 10²⁵ molecules e. 5. 5.62 × 10²⁴ molecules

Answer 1

Answer: E

How much NH₃ can be produced from the reaction below:

N₂ + 3H₂ - 2NH₃

The stoichiometric ratio of the reactants = 1:3

Given

74.2g of N₂, and Molar mass = 14g/mole

Mole of N₂ = 74.2/14=5.3mols of N₂,

and 14mols of H₂

From this given values and comparing with the stoichiometric ratio, H₂ will be the limiting reagent while N₂ is the excess reactant.  

i.e, for every 14mols of H₂, we need 4.67mols of N₂ to react with it to produce 9.33mols of NH₃ as shown (vice versa)

From this we have 9.33mols of NH₃ produced

Avogadro constant, we have n = no of particles = 6.022x10²³ molecules contained in every mole of an element.

For a 9.33mols of NH3, we have 9.33x6.022x10²³molecules in NH3

5.62x10²⁴molecules of NH₃