Question

A yield of NH3 of approximately 98% can be obtained at 200°C and 1,000 atmospheres of pressure.

How many grams of N2 must react to form 1.7 grams of ammonia, NH3?

Answer 1

Answer:

1.429 g of  N₂

Explanation:

The Haber process is a reaction that combines nitrogen with hydrogen to form ammonia according to the following balanced equation:

• N₂ ₍g₎ + 3 H₂ ₍g₎ ⇆  2NH₃ ₍g₎  

One can note that 1 mol of N₂ react with H₂ to produce 2 mol of NH₃.

We cannot compare weight of a substance (in grams) to another in chemical reactions, but we can use moles, then we have to convert the weight of NH3 to moles.

no. of moles of NH₃ = (mass / molar mass) = (1.7 g / 17 g/mol) = 0.1 mol

and the actual yield is 98% , then the theoretical number of moles that would be produced are:  

• percent yield = (actual yield / theoretical yield) × 100

98 = (0.1 mol /  theoretical yield) × 100

theoretical no. of moles of NH₃ = (0.1 * 100) /98 = 0.102 mol

using cross multiplication

1 mol of N₂ → 2 mol of NH₃.

?? mol of N₂ → 0.102 mol of NH₃.

no of moles of N₂ = [(1 mol * 0.102 mol) / 2 mol] = 0.051 mol

Last step is to convert the moles back to grams using:

mass = (no of moles of N₂  * molar mass of N₂)

         = (0.051 mol * 28 g/mol) = 1.429 g