Question

The largest use of nitric acid is in the reaction with ammonia to produce ammonium nitrate for fertilizer.

A) If 275.0 g of nitric acid reacted with 156.5 g of ammonia, what is the limiting reagent?

B) How many grams of ammonium nitrate would be produced?

C) If only 300 g of ammonium nitrate was produced, what is the percent yield?

Answer 1

The largest use of nitric acid is in the reaction with ammonia to produce ammonium nitrate for fertilizer.

1. If 275.0 g of nitric acid reacted with 156.5 g of ammonia, what is the limiting reagent?

Given:

Mass of Nitric acid = 275.0 g , Mass of ammonium = 156.5 g

Solution :

Reaction of HNO3 with ammonia

HNO3 + NH3 -> NH4NO3

To find limiting reactant we use mass of HNO3 and NH3 and to get mol

mol HNO3 = Mass in g / Molar mass = 275.0 g / 63.0119 g mol^-1

= 4.36 mol HNO3

Mol NH3 = 156.5 g / 17.0307 g per mol = 9.19 mol NH3

We use balanced equation to get mol ratio :

1 mol NH3 : 1 mol NH4NO3

1 mol HNO3 : 1 mol NH4NO3

Lets calculate moles of NH4NO3

1. From mol of HNO3

= 4.36 mol HNO3 * 1 mol NH4NO3 / 1mol HNO3

= 4.36 mol NH4NO3

1. from mol NH3

= 9.19 mol NH3 * 1mol NH4NO4 / 1 mol NH3

= 9.19 mol NH4NO3

HNO3 gives less mol of NH4NO3 than NH3 so HNO3 is limiting reactant .

1. How many grams of ammonium nitrate would be produced?

Grams of NH4NO3 produced

Maximum moles of NH4NO3 are formed from limiting reactant

And that is 4.36 mol NH4NO3

Mass of NH4NO3 = mol NH4NO3* Molar mass of NH4NO3

= 4.36 mol * 80.0426 g per mol

= 348.99 g NH4NO3

C) If only 300 g of ammonium nitrate was produced, what is the percent yield?

Percent yield = actual yield / theoretical yield * 100

Actual yield = 300 g

Theoretical yield = Mass of NH4NO3 formed from limiting reactant = 348.99 g

Percent yield = 300/348 * 100

= 86.0 %