Stoichiometry allows us to know the relationship between the amounts of compounds that participate in a chemical reaction. If we look at the stoichiometric coefficients of the adjusted for the synthesis of ammonia, \(N_2 + 3H_2 \rightarrow 2NH_3\), we can deduce that one molecule (or mol) of nitrogen reacts with three of hydrogen to form two of ammonia.

How many grams of ammonia will be obtained from 5.00 grams of nitrogen?

\begin{equation}\bbox[5px,border:2px solid red]{\color{red}{5.00\; g\; N_2\cdot \frac{1\;mol\; N_2}{ 28.01\;g\;N_2}\cdot \frac{2\;mol\;NH_3}{1\;mol\;N_2}\cdot\frac{17.00\;g\;NH_3} {1\;mol\;NH_3}=6.07\;g\;NH_3}}\end{equation}

Let's look at the hydrogen combustion reaction as an example.

\begin{equation} 2H_2(g)+O_2(g)\rightarrow H_2O(l) \end{equation}

The coefficients of the equation (stoichiometric coefficients) allow to know the ratios in moles in which the different substances participate in the reaction. Thus, we can make the following statements:
  • For every two moles of hydrogen that react, two moles of water are produced.
  • Each mole of oxygen consumed produces two moles of water.
  • Two moles of hydrogen react with one mole of oxygen.
  • Example 1. How many moles of water are produced in the above reaction when 2.72 moles of hydrogen are burned in excess of oxygen?

The stoichiometry of the reaction tells us that for every two moles of hydrogen that react, two moles of water are produced. We can write the following conversion factors. \begin{equation} 2.72\;mol\;H_2Ox\frac{2\;mol\;H_2O}{2\;mol\;H_2}=2.72\;mol\;H_2O \end{equation}

Example 2. What mass of water is produced by reacting 4.16 g of $H_2$ with excess oxygen?

To solve this exercise we will follow the following strategy: (1) convert grams to moles; (2) use stoichiometry to calculate the moles of water produced; (3) convert the moles of water to grams.

\begin{equation} 4.16\;g\;H_2x\frac{1\;mol\;H_2}{2.016\;g\;H_2}x\frac{2\;mol\;H_2O}{2\;mol\ ;H_2}x\frac{18.02\;g\;H_2O}{1\;mol\;H_2O}=37.2\;g\;H_2O \end{equation} Example 3. What mass of oxygen is consumed in combustion of 8.68 g of hydrogen?

The steps to follow are: (1) convert grams of hydrogen to moles; (2) stoichiometry tells us that 2 moles of hydrogen react with 1 mole of oxygen; (3) convert moles of oxygen to mass. \begin{equation} 6.86\;g\;H_2x\frac{1\;mol\;H_2}{2.016\;g\;H_2}x\frac{1\;mol\;O_2}{2\;mol\ ;H_2}x\frac{32.00\;g\;O_2}{1\;mol\;O_2}=54.4\;g\;O_2 \end{equation} In the conversion factors we can also include volume, density and composition percentage.

Example 4 . Let the reaction be $2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)$. Determine the volume of the HCl solution (28\% by mass and density 1.14 g/mL) needed to completely react with 1.87 g of Al.

The stages for solving the problem are the following: (1) Convert the grams of Al to moles; (2) use the stoichiometry of the reaction, knowing that 2 moles of aluminum react with 6 moles of HCl, to obtain the required HCl; (3) convert the moles of HCl to grams; (4) use percent by weight to obtain grams of HCl solution; (5) with density convert mass to volume.
\begin{equation} 1.87\;g\;Alx\frac{1\;mol\;Al}{26.98\;g\;Al}x\frac{6\;mol\;HCl}{2\;mol\ ;Al}x\frac{36.46\;g\;HCl}{1\;mol\;HCl}x\frac{100\ g\ dis.\ HCl}{28\ g\ HCl}x\frac{1\ mL\ dis.\ HCl}{1.14\ g\ dis.\ HCl}=23.8\ mL\ dis.\ HCl \end{equation}