Page 145 - Microsoft Word - LessonPlan-Overview.doc

Answers to Advanced Chemistry Exercises

1. Determine the number of moles of N 2O 4 needed to react completely
with 2.56 mol of N 2H 4 for the reaction 2 N 2H 4(l) + N 2O 4(l) → 3 N 2(g) +
4 H 2O(l)

Find the relation between moles of N 2H 4 and N 2O 4 by using the
coefficients of the balanced equation: 2 mol N 2H 4 is proportional to 1
mol N 2O 4 Therefore, the conversion factor is 1 mol N 2O 4/2 mol N 2H 4:

moles N 2O 4 = 2.56 mol N 2H 4 x 1 mol N 2O 4/2 mol N 2H 4

moles N 2O 4 = 1.28 mol N 2O 4 (answer)

2. Determine the number of moles of N 2 produced for the reaction 2
N 2H 4(l) + N 2O 4(l) → 3 N 2(g) + 4 H 2O(l) when the reaction begins with
4.52 moles of N 2H 4.

Find the relation between moles of N 2H 4 and N 2 by using the
coefficients of the balanced equation: 2 mol N 2H 4 is proportional to 3
mol N 2 In this case, we want to go from moles of N 2H 4 to moles of N 2,
so the conversion factor is 3 mol N 2/2 mol N 2H 4:

moles N 2 = 4.52 mol N 2H 4 x 3 mol N 2/2 mol N 2H 4

moles N 2 = 6.78 mol N 2O 4 (answer)

3. The balanced equation for the synthesis of ammonia is 3 H 2(g) +
N 2(g) → 2 NH 3(g). 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 72.0 g of N 2:

mass NH 3 = 72.0 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 = 87.43 g NH 3 (answer)

© 2010 Supercharged Science www.ScienceLearningSpace.com

547