**Problem**

The balanced equation for the synthesis of ammonia is 3 H

_{2}(g) + N

_{2}(g) → 2 NH

_{3}(g).

Calculate:

a. the mass in grams of NH

_{3}formed from the reaction of 64.0 g of N

_{2}

b. the mass in grams of N

_{2}required for form 1.00 kg of NH

_{3}

**Solution**

From the balanced equation, it is known that:

1 mol N

_{2}∝ 2 mol NH

_{3}

Use the periodic table to look of the atomic weights of the elements to calculate the weights of the reactants and products:

1 mol of N

_{2}= 2(14.0 g) = 28.0 g

1 mol of NH

_{3}is 14.0 g + 3(1.0 g) = 17.0 g

These relations can be combined to give the conversion factors needed to calculate the mass in grams of NH

_{3}formed from 64.0 g of N

_{2}:

mass NH

_{3}= 64.0 g N

_{2}x 1 mol N

_{2}/28.0 g NH

_{2}x 2 mol NH

_{3}/1mol NH

_{3}x 17.0 g NH

_{3}/1 mol NH

_{3}

mass NH

_{3}= 77.7 g NH

_{3}

To obtain the answer to the second part of the problem, the same conversions are used, in a series of three steps:

(1) grams NH

_{3}→ moles NH

_{3}(1 mol NH

_{3}= 17.0 g NH

_{3})

(2) moles NH

_{3}→ moles N

_{2}(1 mol N

_{2}∝ 2 mol NH

_{3})

(3) moles N

_{2}→ grams N

_{2}(1 mol N

_{2}= 28.0 g N

_{2})

mass N

_{2}= 1.00 x 10

^{3}g NH

_{3}x 1 mol NH

_{3}/17.0 g NH

_{3}x 1 mol N

_{2}/2 mol NH

_{3}x 28.0 g N

_{2}/1 mol N

_{2}

mass N

_{2}= 824 g N

_{2}

**Answer**

a. mass NH

_{3}= 77.7 g NH

_{3}

b. mass N

_{2}= 824 g N

_{2}