4.3 Reaction Stoichiometry

Reaction Stoichiometry

Reaction stoichiometry lets you predict exactly how much of each substance is consumed or produced in a chemical reaction. It's the math that connects a balanced equation on paper to real quantities you can measure in the lab.

This section covers how to use mole ratios from balanced equations to calculate masses, volumes, and concentrations of reactants and products.

Stoichiometry in Chemical Reactions

Stoichiometry is the quantitative study of reactants and products in a chemical reaction. It answers questions like: How many grams of oxygen do you need to burn 10 grams of hydrogen?

The whole approach rests on the law of conservation of mass: atoms are neither created nor destroyed in a reaction, just rearranged. That means the total mass of reactants always equals the total mass of products.

The tool that makes stoichiometry work is the mole ratio, which comes directly from the coefficients in a balanced equation. For example, in:

$H_2 + Cl_2 \rightarrow 2\,HCl$

the coefficients tell you that 1 mole of $H_2$ reacts with 1 mole of $Cl_2$ to produce 2 moles of $HCl$. Those coefficients are the mole ratio (1:1:2), and every stoichiometry calculation depends on them.

Reactant and Product Quantity Calculations

Every stoichiometry problem follows the same general process. Here are the steps:

1. Write and balance the chemical equation. Make sure each element has the same number of atoms on both sides. For example: $2\,H_2 + O_2 \rightarrow 2\,H_2O$

2. Identify the mole ratio. Read the coefficients. In the equation above, the ratio is 2 mol $H_2$ : 1 mol $O_2$ : 2 mol $H_2O$.

3. Convert your known quantity to moles. Whatever you're given (mass, volume, etc.), get it into moles first.

   • For mass: divide by molar mass. Example: $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$
   • For gases at STP: divide by molar volume (22.4 L/mol). Example: $\text{moles} = \frac{\text{volume (L)}}{22.4\,\text{L/mol}}$
   • For solutions: multiply molarity by volume in liters. Example: $\text{moles} = M \times V$

4. Use the mole ratio to find moles of the unknown. Set up a conversion using the ratio from step 2. If you have 4 mol $H_2$ and need mol $O_2$: $4\,\text{mol}\,H_2 \times \frac{1\,\text{mol}\,O_2}{2\,\text{mol}\,H_2} = 2\,\text{mol}\,O_2$

5. Convert moles of the unknown to the unit you need (grams, liters, etc.) using the reverse of step 3.

This five-step process works for virtually every stoichiometry problem you'll encounter. The key is always going through moles as the bridge between substances.

Mass, Mole, and Concentration Problems

The steps above apply to all stoichiometry problems, but the conversions in steps 3 and 5 change depending on what units you're working with. Here's how each type works:

Mass problems use molar mass to convert between grams and moles.

• Grams to moles: $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$
  • Example: $\frac{5\,\text{g NaCl}}{58.44\,\text{g/mol}} = 0.0856\,\text{mol NaCl}$
• Moles to grams: $\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$

Gas problems at STP use molar volume (22.4 L/mol at 0°C and 1 atm).

• Example: $1\,\text{mol}\,O_2 \times 22.4\,\text{L/mol} = 22.4\,\text{L}\,O_2$

Solution problems use molarity (M), which is moles of solute per liter of solution.

• $M = \frac{\text{moles of solute}}{\text{liters of solution}}$
• To find moles from a solution: $\text{moles} = M \times V$
  • Example: $0.5\,M \times 2.5\,\text{L} = 1.25\,\text{mol NaCl}$
• To find volume needed: $V = \frac{\text{moles}}{M}$
  • Example: $\frac{0.5\,\text{mol NaCl}}{0.5\,M} = 1\,\text{L solution}$

Dimensional analysis is your best friend here. Write out your units at every step and cancel them as you go. If the units don't work out, something's off in your setup.

Additional Stoichiometric Concepts

A few related ideas show up alongside stoichiometry problems:

• Theoretical yield is the maximum amount of product you'd get if the reaction went perfectly, calculated using stoichiometry. Percent yield compares what you actually collected to this theoretical maximum: $\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$
• Limiting reactant is the reactant that runs out first, which determines how much product can form. The other reactant left over is the excess reactant.
• Stoichiometric proportions describe the condition where reactants are present in exactly the ratio the balanced equation requires, with nothing left over.