Calculate stoichiometric reacting masses, limiting reactants, volumes of gases at r.t.p., volumes of solutions and concentrations of solutions expressed in $mathrm{g} / mathrm{dm}^3$ and $mathrm{mol} / mathrm{dm}^3$, including conversion between $mat

Stoichiometry – The Mole and the Avogadro Constant

1️⃣ What is a Mole?

A mole is a counting unit – just like a dozen is 12 eggs. One mole contains exactly \$6.022\times10^{23}\$ entities (atoms, molecules, ions, etc.). This number is called the Avogadro constant.

Think of it as a “super‑counting” tool: if you have 1 mol of water, you have 6.022 × 10²³ water molecules – that’s a lot of tiny droplets!

2️⃣ Avogadro Constant in Action

The Avogadro constant links mass and number of entities:

\$\$

\text{Number of entities} = \frac{\text{Mass (g)}}{\text{Molar mass (g mol}^{-1}\text{)}} \times N_A

\$\$

where \$N_A = 6.022\times10^{23}\ \text{mol}^{-1}\$.

3️⃣ Calculating Stoichiometric Masses

Given a balanced equation, you can convert between masses of reactants and products using molar masses and mole ratios.

4️⃣ Limiting Reactant – The “Race” Analogy 🚦

Imagine two runners (reactants) in a race to produce a product. The runner who finishes first (uses up all its material) determines how much product can be made – that’s the limiting reactant.

1. Calculate moles of each reactant.
2. Divide by the stoichiometric coefficient to get the available moles of product per reactant.
3. The smallest value indicates the limiting reactant.
4. Use that value to find the maximum product mass.

5️⃣ Volumes of Gases at R.T.P. (25 °C, 1 atm)

At R.T.P., 1 mol of any ideal gas occupies \$22.4\ \text{dm}^3\$ (or 22 400 cm³). Therefore:

\$\$

V = n \times 22.4\ \text{dm}^3

\$\$

If you know the volume, you can find moles:

\$\$

n = \frac{V}{22.4\ \text{dm}^3}

\$\$

6️⃣ Volumes of Solutions & Concentrations

- Concentration (C) is mass of solute per unit volume of solution:

\$\$

C{\text{(g dm}^{-3}\text{)}} = \frac{m{\text{solute}}}{V_{\text{solution}}}

\$\$

- Molarity (M) is moles of solute per litre (dm³) of solution:

\$\$

M = \frac{n{\text{solute}}}{V{\text{solution}}}

\$\$

Remember: 1 dm³ = 1000 cm³. So, to convert:

\$\$

V{\text{dm}^3} = \frac{V{\text{cm}^3}}{1000}

\$\$

and vice‑versa.

7️⃣ Quick Reference Table 📚

8️⃣ Example Problem 🧪

Problem: 10 g of sodium (Na) reacts with 20 g of chlorine gas (Cl₂) to form sodium chloride (NaCl). Which reactant is limiting? How many grams of NaCl are produced?

1. Find moles:

\$\$

n_{\text{Na}} = \frac{10}{22.99} = 0.435\ \text{mol}

\$\$

\$\$

n{\text{Cl}2} = \frac{20}{70.90} = 0.282\ \text{mol}

\$\$

2. Balanced equation: \$2\,\text{Na} + \text{Cl}_2 \rightarrow 2\,\text{NaCl}\$

3. Check limiting reactant:

- Na needed per Cl₂: 2 mol Na / 1 mol Cl₂ → 0.282 mol Cl₂ × 2 = 0.564 mol Na required.

- We only have 0.435 mol Na → Na is limiting.

4. Maximum NaCl produced:

\$\$

n{\text{NaCl}} = n{\text{Na}} = 0.435\ \text{mol}

\$\$

\$\$

m_{\text{NaCl}} = 0.435 \times 58.44 = 25.4\ \text{g}

\$\$

So, 25.4 g of NaCl can be made from the given amounts.

9️⃣ Tips for Success

• Always write the balanced equation first.
• Use the mole ratio directly from the coefficients.
• Keep units consistent – grams, moles, dm³, cm³.
• Check your work by verifying that the limiting reactant is indeed the smallest amount of product you can produce.
• Remember: 1 dm³ = 1000 cm³.

🔚 Final Thought

Stoichiometry is like a recipe: you need the right amounts of each ingredient to get the perfect dish. With the mole, Avogadro constant, and a few conversion tricks, you can calculate exactly how much of every substance you need or will produce. Happy calculating! 🚀