Answer to Question #263222 in Inorganic Chemistry for gela

Question

Provide the correct answer. Submit your solution in the gclass. Consider the reaction below. If 10.10 moles H2 are made to react with 8.64 moles N2, a. what is the limiting reagent? b. How many grams of NH3 will be produced? c. What reagent is in excess and by how much? d. Compute for its percent yield if 78.09 g was produced. See answer format. 

Answer 1

N₂ + 3H₂ → 2NH₃

According to the reaction equation for 1 mol of N₂ are needs 3 moles of H₂.

Therefore, for 8.64 moles N₂ are needs 25.92 moles H₂.

The limiting reagent is H₂.

The theoretical yield of NH3 is $2×8.64=17.28moles$ .

Mass of $NH_3= 17.031 ×17.28=294.3grams$