What is the maximum amount of NH3 gas formed by mixing of 14 g N2 and 6 g H2 gases is?

Answer 1

N2(g) + 3H2(g) ==> 2NH3(g) is the balanced equation for this reaction.14 g N2 x 1 mole N2/28 g = 0.5 moles N2 present

6 g H2 x 1 mole H2/2 g = 3 moles O2 present

In this situation, the amount of N2 is limiting the overall reaction. So, the maximum amount of NH3 that can be produced will be limited by the amount of N2.

0.5 moles N2 x 2 moles NH3/1 mole N2 = 1 moles NH3 is the maximum amount of NH3.

Answer 2

The maximum amount ammonia that can be produced is 17 g.

This is a limiting reagent (reactant) question. You will need to determine the mass of NH3 produced by each reactant.

Start with a balanced equation.

N2(g) + 3H2(g) --> 2NH3(g)

You need to determine the molar masses of each gas. I'm going to round to whole numbers.

N2: 2 x 14 g/mol N = 28 g/mol N2

H2: 2 x 1 g/mol H = 2 g/mol H2

NH3: (1 x 14 g/mol N) + (3 x 1 g/mol H) = 17 g/mol NH3

The process to use is:

Determine the moles of each reactant from their given masses by dividing their given masses by their molar masses. Since molar mass (g/mol) is a fraction, you can divide by multiplying the given mass by the reciprocal of the molar mass (mol/g).

Determine the moles of ammonia produced by multiplying the moles of each reactant by the mole ratio between it and ammonia from the balanced equation. Ammonia will go in the numerator. Then multiply the moles of ammonia by its molar mass.

Mass of NH3 produced by 14 g N2

14 g N2 x (1 mol N2)/(28 g N2)/(1 mol N2) x (2 mol NH3)/(1 mol N2) x (17.031 g NH3)/(1 mol NH3) =17 g NH3 rounded to two significant figures

Mass of NH3 produced by 6 g H2

6 g H2 x (1 mol H2)/(2 g H2) x (2 mol NH3)/(3 mol H2) x (17.031 g NH3)/(1mol NH3) = 34 g NH3 rounded to one significant figure

Since N2 produces less NH3, it is the limiting reagent and this reaction can produce a maximum of 17 g NH3.