Extended Reaction Stoichiometry Road Map — Examples

Stoichiometry Road Map

Below is a road map to help figure out how to do some conversion commonly done in chemistry

Mole Ratio: the mole ratio comes from the chemical equation

• The coefficients of the chemical equation tell you the mole ratio between each reactant and product in the chemical equation

Note: this map shows you how to convert in one direction (from moles to mass or from moles to volume), in order to go from volume to moles or mass to moles you simply just multiply by the reciprocal

• Ex: Volume → Moles : Volume $\cdot\left(\frac{1mol}{22.4L}\right)$ = Moles

We've been studying chemical reactions. After we learned the different chemical reactions, we started looking at stoichiometry. Stoichiometry helps us determine the amount of product we produce. By now, you're starting to get the hang of stoichiometry. You've seen all the different conversions you can do. Here is a road map that summarizes those conversions. It is a tool that guides you through solving stoichiometry problems.

Stoichiometry Roadmap

$$\begin{array}{c|c|c} \textbf{Start With...} & \textbf{Conversion} & \textbf{End With...} \\ \hline Gas\,Volume\,at\,STP & multiply\,by\,22.4(\frac{L}{mol}) & Moles\,of\,Gas \\ \hline Mass & Divide\,by\,Molar\,Mass\,(\frac{g}{mol}) & Moles \\ \hline Particles & divide\,by\,6.023 \times 10^{23}(\frac{particles}{mol}) & Moles \\ \hline Concentration\,(\frac{mol}{L}) & Multiply\,by\,Volume\,(L) & Moles \\ \hline Volume (L) & Multiply\,by\,Density (\frac{g}{L}) & Mass \\ \hline Moles\,of\,Reactant & Use\,Mole\,Ratio & Moles\,of\,Product \end{array}$$

The roadmap can help you convert...

• moles of reactant to moles of product.

• mass of reactant to mass of product.

• the volume of reactant to the volume of product. To find the liquid's volume, you will need to do the stochiometry with density.

Steps in a Stoichiometry Problem

Let's review the steps in a stoichiometry problem.

Step 1: Write the chemical equation.

Step 2: Balance the chemical equation.

Step 3: Do any needed conversions. You need SI units.

Step 4: Find the moles of the reactant(s).

• Use the stoichiometry road map.

• Gas stoichiometry problems may require you to use the ideal gas law.

Step 5: Determine the limiting reagent.

Step 6: Find the moles of the product(s). Use the mole ratios.

Step 7: If the question asks for something other than moles of product, solve the question.

• If the product is a gas, you may need to solve for pressure, volume, or temperature. I recommend using the ideal gas law.

• Remember, at STP, one mole of gas is 22.4 liters.

• If the product is a solid, you may have to find its mass.

• If the product is a liquid, you may have to find its mass or volume. Converting from one to the other requires its density.

• If the product is aqueous, you will need to find the concentration.

Example Problem

Solid iron(II) sulfide undergoes a double displacement reaction with hydrochloric acid. The reaction produces solid iron(II) chloride and gaseous hydrogen sulfide. You start with 90.0 grams of iron(II) sulfide. Next, you react it with 1425.0 milliliters of 1.50 molar hydrochloric acid. You are at STP. Calculate the mass of iron(II) chloride, the volume of hydrogen sulfide, and the particles of hydrogen sulfide at the end of the reaction.

Step 1: Write the chemical equation.

• $FeS_{(s)} + HCl_{(aq)} \rightarrow FeCl_{2(s)} + H_{2}S_{(g)}$

• If you had trouble with the names, look at naming ionic compounds, naming covalent compounds, and naming acids and bases.

Step 2: Balance the chemical equation.

• $FeS_{(s)} +2 HCl_{(aq)} \rightarrow FeCl_{2(s)} + H_{2}S_{(g)}$

Step 3: Do any necessary conversions.

• The only conversion we need to do is to convert milliliters to liters.

• $1425.0ml \times \frac{1.0 L}{1000.0ml} =1.425 L$

Step 4: Find the moles of reactants.

• Here we use our stoichiometry roadmap.

• We need to convert the mass of iron(II) sulfide and the concentration of hydrochloric acid to moles. Use the periodic table to calculate the molar mass of iron(II) sulfide.

• $\frac{90.0g}{87.91\frac{g}{mol}}= 1.02 mols\,FeS_{(s)}$

• $1.50\frac{mols}{L}\times 1.425L =2.14 mols\,HCl_{(aq)}$

Step 5: Find the limiting reagent.

• To find the limiting reagent use the mole ratios.

• $\frac{1.02mols_{FeS}}{1}= 1.02 mols$

• $\frac {2.14mols_{HCl}}{2}= 1.07 mols$

• So, iron(II) sulfide will be the limiting reagent.

Step 6: Find the moles of the products.

• Again we use the mole ratios.

• $\frac{1 mol FeCl_{2(s)}}{2 mols HCl_{(aq)}}\times 2.14mols HCL=1.07\, mols\,FeCl_{2(s)}$

• $\frac{1 mol H_{2}S_{(g)}}{2 mols HCl_{(aq)}}\times 2.14mols HCL=1.07 mols\,H_{2}S_{(g)}$

Step 7: Answer the problem.

• Again, we use the stoichiometry road map.

• $1.07mols_{FeCl_{2(s)}} \times \frac{126.75g}{mol}= 135.62\,g\,FeCl_{2(s)}$

• $1.07 moles_{H_{2}S_{(g)}} \times \frac{22.4L}{mol}= 23.4\, L\,H_{2}S_{(g)}$

• $1.07 moles_{H_{2}S_{(g)}}\times\frac{6.023\times10^{23} particles}{mol}= 6.44 \times10^{23}\, particles\,H_{2}S_{(g)}$

(Video) Stoichiometry Road Map

By Hannah Nandor

This short video walks you step-by-step through the stoichiometry road map. If you're struggling with stoichiometry problems, this video can help!