Stoichiometry: Videos & Practice Problems

Balancing chemical equations is a crucial first step in understanding stoichiometry, which focuses on the quantitative relationships between reactants and products in a chemical reaction. Stoichiometry enables us to calculate the amounts of products formed from given reactants and vice versa. For instance, consider the balanced chemical equation:

2 H₂ (g) + O₂ (g) → 2 H₂O (g)

In this example, 12.3 grams of hydrogen gas (H₂) are provided, and the goal is to determine the mass of water (H₂O) produced. This process exemplifies stoichiometry, where we use the balanced equation to relate the quantities of different substances involved in the reaction.

To perform stoichiometric calculations, one must follow specific procedures that involve converting grams to moles, using the mole ratio from the balanced equation, and then converting back to grams if necessary. Understanding these steps is essential for accurately determining the amounts of reactants and products in any chemical reaction.

To determine how many grams of water (H₂O) can be formed from 12.3 grams of hydrogen gas (H₂), we utilize the principles of stoichiometry. Stoichiometry allows us to convert between the quantities of reactants and products in a chemical reaction using a stoichiometric chart.

The first step involves converting the given mass of hydrogen gas into moles. This is essential because stoichiometric calculations are based on the mole concept. The conversion from grams to moles is done using the molar mass of hydrogen, which is approximately 2.02 g/mol. Thus, the number of moles of H₂ can be calculated as:

\[\text{Moles of } H_2 = \frac{\text{grams of } H_2}{\text{molar mass of } H_2} = \frac{12.3 \text{ g}}{2.02 \text{ g/mol}} \approx 6.09 \text{ moles}\]

Next, we need to perform a mole-to-mole comparison using the coefficients from the balanced chemical equation for the reaction between hydrogen and oxygen to form water. The balanced equation is:

\[2H_2 + O_2 \rightarrow 2H_2O\]

This equation indicates that 2 moles of hydrogen gas produce 2 moles of water. Therefore, the mole ratio of H₂ to H₂O is 1:1. This means that the moles of water produced will be equal to the moles of hydrogen consumed:

\[\text{Moles of } H_2O = 6.09 \text{ moles of } H_2 \times \frac{2 \text{ moles of } H_2O}{2 \text{ moles of } H_2} = 6.09 \text{ moles of } H_2O\]

Finally, to find the mass of water produced, we convert moles of water back to grams using the molar mass of water, which is approximately 18.02 g/mol:

\[\text{Grams of } H_2O = \text{Moles of } H_2O \times \text{molar mass of } H_2O = 6.09 \text{ moles} \times 18.02 \text{ g/mol} \approx 109.1 \text{ g}\]

In summary, starting from the given mass of hydrogen, we successfully navigated through the stoichiometric chart, converting grams to moles, performing a mole-to-mole comparison, and finally converting back to grams to find that approximately 109.1 grams of water can be produced from 12.3 grams of hydrogen gas.

In stoichiometry, we often need to convert between the amounts of different substances in a chemical reaction. This process involves using a balanced chemical equation to relate the quantities of reactants and products. For example, if we start with 12.3 grams of hydrogen gas (H₂), we can determine how many grams of water (H₂O) are produced.

To solve this problem, we follow a systematic approach. First, we convert the given mass of H₂ into moles. The molar mass of H₂ is calculated as follows: each hydrogen atom has a mass of approximately 1.008 grams, so for H₂, the molar mass is:

Molar mass of H₂ = 2 × 1.008 g/mol = 2.016 g/mol

Using this, we convert grams to moles:

Moles of H₂ = \(\frac{12.3 \text{ g H}_2}{2.016 \text{ g/mol}} \approx 6.1012 \text{ moles H}_2\)

Next, we use the stoichiometric coefficients from the balanced equation to convert moles of H₂ to moles of H₂O. The balanced equation indicates that 2 moles of H₂ produce 2 moles of H₂O, allowing us to set up the conversion:

Moles of H₂O = 6.1012 moles H₂ × \(\frac{2 \text{ moles H}_2\text{O}}{2 \text{ moles H}_2}\) = 6.1012 moles H₂O

Finally, we convert moles of H₂O to grams. The molar mass of water is calculated as follows: it consists of 2 hydrogen atoms and 1 oxygen atom:

Molar mass of H₂O = (2 × 1.008 g) + (16.00 g) = 18.016 g/mol

Now, we can convert moles of H₂O to grams:

Grams of H₂O = 6.1012 moles H₂O × 18.016 g/mol ≈ 109.9 g H₂O

Considering significant figures, we round our final answer to three significant figures, resulting in:

Final answer: 1.10 × 10² g H₂O

This example illustrates the importance of stoichiometry in converting between different substances in a chemical reaction, emphasizing the need for a balanced equation and careful unit conversions.