Chemical Equations & Stoichiometry

25min Part 1 / Ch2 / Lesson 3

Prerequisites: 1-2-1

Objectives

Write and balance chemical equations
Use coefficient ratios to determine mole relationships
Perform stoichiometric calculations

Writing Chemical Equations

Three rules for chemical equations:

Reactants on the left, products on the right
Balance atoms on both sides (law of conservation of mass)
Use the simplest whole-number coefficients

Example: Combustion of hydrogen

2\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\mathrm{H_2O}

Atom	Left side	Right side
H	$2 \times 2 = 4$	$2 \times 2 = 4$ ✓
O	$1 \times 2 = 2$	$2 \times 1 = 2$ ✓

Balancing by Inspection

Example: Complete combustion of ethanol

Balance: $\mathrm{C_2H_5OH + O_2 \rightarrow CO_2 + H_2O}$

$a\mathrm{C_2H_5OH} + b\mathrm{O_2} \rightarrow c\mathrm{CO_2} + d\mathrm{H_2O}$

C: $2a = c$
H: $6a = 2d$ → $d = 3a$
O: $a + 2b = 2c + d$

$a=1$ → $c=2$ , $d=3$

O: $1 + 2b = 2(2) + 3 = 7$ → $b = 3$

$\mathrm{C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O}$

Coefficients = Mole Ratios

Coefficients represent mole ratios

Coefficient ratio = mole ratio. This is the foundation of all stoichiometric calculations.

Problem-Solving Strategy

Stoichiometry calculation flowchart

Worked Example

Example: Combustion of magnesium

How many grams of MgO are produced when $4.8\ \mathrm{g}$ of Mg is completely burned? ( $\mathrm{Mg} = 24$ , $\mathrm{O} = 16$ )

$2\mathrm{Mg} + \mathrm{O_2} \rightarrow 2\mathrm{MgO}$

$n_{\mathrm{Mg}} = \frac{4.8}{24} = 0.20\ \mathrm{mol}$

Mg : MgO = 2 : 2 = 1 : 1, so

$n_{\mathrm{MgO}} = 0.20\ \mathrm{mol}$

$m = 0.20 \times (24 + 16) = 0.20 \times 40 = 8.0\ \mathrm{g}$

Check Your Understanding

Q1 In 2H₂ + O₂ → 2H₂O, how much O₂ is needed to react with 4 mol H₂?

Q2 In N₂ + 3H₂ → 2NH₃, how much N₂ is needed to produce 0.60 mol NH₃?

Q3 CaCO₃ → CaO + CO₂: What volume of CO₂ (at STP) is released from 10 g CaCO₃? (CaCO₃ = 100)

Exercises

Q1. In the reaction $\mathrm{Fe + 2HCl \rightarrow FeCl_2 + H_2}$ , find the volume of hydrogen gas (at STP) produced from $5.6\ \mathrm{g}$ of Fe reacting with excess HCl. ( $\mathrm{Fe} = 56$ )

Solution

$n_{\mathrm{Fe}} = \frac{5.6}{56} = 0.10\ \mathrm{mol}$

Fe : H₂ = 1 : 1, so $n_{\mathrm{H_2}} = 0.10\ \mathrm{mol}$

$V = 0.10 \times 22.4 = 2.24\ \mathrm{L}$

Q2. Write the balanced equation for the complete combustion of propane ( $\mathrm{C_3H_8}$ ). Then calculate the masses of $\mathrm{CO_2}$ and $\mathrm{H_2O}$ produced from burning $11\ \mathrm{g}$ of propane. ( $\mathrm{C}=12$ , $\mathrm{H}=1$ , $\mathrm{O}=16$ )

Solution

$\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$

$n_{\mathrm{C_3H_8}} = \frac{11}{44} = 0.25\ \mathrm{mol}$

$n_{\mathrm{CO_2}} = 0.25 \times 3 = 0.75\ \mathrm{mol}$ → $m = 0.75 \times 44 = 33\ \mathrm{g}$

$n_{\mathrm{H_2O}} = 0.25 \times 4 = 1.0\ \mathrm{mol}$ → $m = 1.0 \times 18 = 18\ \mathrm{g}$