Chapter 1. Diagnostic Quiz 4.2

Introduction

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You must read each slide, and complete any questions on the slide, in sequence.

4.2 Balancing chemical reaction equations

Solved Problem 4.2-1 Balancing a chemical reaction equation

Balance the following chemical equation: NaBH₄ + BF₃ -> NaBF₄ + B₂H₄

Strategy

Inspect the equation and compare the relative numbers of atoms in reactants and products. Then place coefficients in front of reactant or product species to give equal numbers of all atoms (on both sides of the arrow). It is helpful to begin with the atom that appears the fewest times on each side of the equation.

Solution

Counting the number of atoms of each type of element in reactants and product we find:

NaBH₄ + BF₃-> NaBF₄ + B₂H₄(unbalanced)

Count:	1 Na atom	1Na atom
	2 B atoms	3 B atoms
	3 F atoms	4 F atoms
	4 H atoms	4 H atoms

In this case, sodium and hydrogen atoms are balanced; boron and fluorine are not. Notice that B appears in two formulae on each side while F appears in only one formula on each side of the equation. So we’ll start by balancing the F atoms. To balance fluorine, we first find the lowest common multiple of 3 and 4 which is 12. So we use coefficients that give 12 F atoms on both sides by entering the coefficient 4 for BF₃ and 3 for NaBF₄:

NaBH₄ + 4BF₃-> 3NaBF₄ + B₂H₄(unbalanced)

Next, look at boron. Notice that the change in coefficients now produces 5 boron atoms on the left and 5 boron atoms on the right. The equation is balanced.

Solved- Problem 4.2-2 Balancing a chemical reaction equation

Balance the following chemical equation: P₄O₆ + I₂ -> P₂I₄ + P₄O₁₀

Strategy

Solution

Counting the number of atoms of each type of element in reactants and product we find:

P₄O₆ + I₂-> P₂I₄ + P₄O₁₀ (unbalanced)

Count:	4 P atoms	6 P atoms
	6 O atoms	10 O atoms
	2 I atoms	4 I atoms

Both oxygen and iodine appear in only one species on each side of the equation. However, since iodine appears on the reactant side only as a diatomic molecule, we can balance that last, since changing its coefficient will only affect one atom type. So let’s start with oxygen. To balance the oxygen atoms we first find the lowest common multiple of 6 and 10, which is 30. These multiples are 5 and 3 so we set up the equation with 30 O atoms on both sides:

5P₄O₆ + I₂-> P₂I₄ + 3P₄O₁₀(unbalanced)

This change leads to 20 P on the left and 14 P on the right; we need 8 more P atoms on the right so we use the coefficient 4 to balance the P atoms:

5P₄O₆ + I₂-> 4P₂I₄ + 3P₄O₁₀(unbalanced)

Finally, we have 2 I atoms on the left and 16 on the right. We use the coefficient 8 to balance the I atoms:

5P₄O₆ + 8I₂-> 4P₂I₄ + 3P₄O₁₀(balanced)

Solved- Problem 4.2-3 Balancing a chemical reaction equation

Balance the following chemical equation: C₆H₁₂O₆ + O₂ -> CO₂ + H₂O

Strategy

Inspect the equation and compare the relative numbers of atoms in reactants and products. Then place coefficients in front of reactant or product species to give equal numbers of all atoms (on both sides of the arrow). It is helpful to begin with an atom that is part of the most complex substance because changing coefficients for simpler substances affects fewer elements.

Solution

Counting the number of atoms of each type of element in reactants and product we find:

C₆H₁₂O₆(s) + O₂(g) -> CO₂(g) + H₂O(l) (unbalanced)

Count:	6 C atoms	1 C atom
	12 H atoms	2 H atom
	8 O atoms	3 O atoms

We start by balancing carbon since it is part of the most complex substance, C₆H₁₂O₆. The coefficient 6 in front of CO₂ balances the C atoms.

C₆H₁₂O₆(s) + O₂(g) -> 6 CO₂(g) + H₂O(l) (unbalanced)

Next, balance hydrogen with the coefficient 6 in front of H₂O:

C₆H₁₂O₆(s) + O₂(g) -> 6 CO₂(g) + 6H₂O(l) (unbalanced)

These two coefficients lead to a total of 18 O on the right side and 8 O on the left. To get 10 more O atoms on the left, we place the coefficient 6 in front of O₂.

C₆H₁₂O₆(s) + 6O₂(g) -> 6 CO₂(g) + 6H₂O(l) (balanced)