Question

What mass of water must be decomposed to produce \(25.0 \mathrm{~L}\) of oxygen at STP?

Short answer

40.2 g of water must be decomposed.

Step-by-step solution

Write the balanced chemical equation

The decomposition of water can be represented by the balanced chemical equation:a) \[\begin{equation}2 \text{H}_2\text{O}(l) \rightarrow 2 \text{H}_2(g) + \text{O}_2(g)\texttt{} \texttt{} \texttt{} \texttt{} \texttt{} \texttt{} \end{equation}\]\texttt{} \texttt{} \texttt{} \texttt{}\texttt{} \texttt{} \texttt{} \texttt{} \texttt{} \texttt{} For every 2 moles of water decomposed, 1 mole of oxygen gas is produced.

Use the ideal gas law to find moles of oxygen (O₂)

At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 L. To find the moles of \text{O_2} produced, divide the volume of \text{O_2} by the molar volume at STP:\[\begin{equation} n=\frac {V}{V_m}= \frac{25.0 L}{22.4 L/mol} =1.116 mol \end{equation}\]

Understand the stoichiometric ratio

According to the balanced chemical equation, 2 moles of H₂O produce 1 mole of O₂. Therefore, for 1.116 moles of O₂:\[\begin{equation} moles of H₂O = 2 × 1.116 mol = 2.232 mol \end{equation}\]

Convert moles of water to mass

The molar mass of water (\text{H}_2\text{O}) is 18.02 g/mol. Thus, the mass of water required can be calculated as:\[\begin{equation} mass = moles × molar mass = 2.232 mol × 18.02 g/mol = 40.2 g \end{equation}\]

Key concepts

chemical equation

In chemistry, a chemical equation is a way to depict a chemical reaction using symbols and formulas. The equation shows the reactants (substances that start the reaction) and the products (substances that are produced in the reaction). An important part of a chemical equation is balancing it. This means ensuring that the number of atoms of each element is the same on both sides of the equation. For instance, when water (H₂O) decomposes, it forms hydrogen gas (H₂) and oxygen gas (O₂). The balanced chemical equation for this reaction is:
\[2 \text{H}_2\text{O} \rightarrow 2 \text{H}_2 + \text{O}_2\]
Here, 2 moles of water decompose to produce 2 moles of hydrogen gas and 1 mole of oxygen gas. Balancing ensures that the principle of conservation of mass is followed.

ideal gas law

The ideal gas law is an equation of state for a hypothetical ideal gas. It provides a good approximation for the behavior of many gases under various conditions. The ideal gas law is written as:
\[PV = nRT\]
Where:

• P is the pressure of the gas
• V is the volume of the gas
• n is the number of moles of the gas
• R is the gas constant (0.0821 L·atm/(mol·K))
• T is the temperature in Kelvin

For example, at standard temperature and pressure (STP, which is 0°C or 273K and 1 atm), 1 mole of any ideal gas occupies 22.4 L. To find the number of moles () of oxygen gas (\text{O}_2) produced from the decomposition of water at STP, we can use this law. If we have 25.0 L of \text{O}_2:
\[n = \frac {V}{V_m} = \frac{25.0 L}{22.4 L/mol} = 1.116 mol\]

stoichiometric ratio

The stoichiometric ratio in a chemical reaction is the molar relationship between the reactants and products. It is determined from the balanced chemical equation. In our example, the balanced equation for the decomposition of water is:
\[2 \text{H}_2\text{O} \rightarrow 2 \text{H}_2 + \text{O}_2\]
This tells us that 2 moles of water produce 2 moles of hydrogen gas and 1 mole of oxygen gas. From this, we can find that the ratio is 2:1 between \text{H}_2\text{O} and \text{O}_2. Therefore, for every mole of \text{O}_2 produced, 2 moles of \text{H}_2\text{O} are required. With 1.116 moles of \text{O}_2 produced, the moles of \text{H}_2\text{O} needed are:
\[2 \times 1.116 mol = 2.232 mol\]

molar mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is a crucial concept in chemistry because it allows us to convert between the mass of a substance and the amount in moles. The molar mass of a compound is calculated by adding the molar masses of its constituent elements. For water (H₂O), the molar masses are:

• Hydrogen (H) = 1.01 g/mol
• Oxygen (O) = 16.00 g/mol

Thus, the molar mass of water is:
\[2 \times 1.01 + 16.00 = 18.02 \text{ g/mol}\]
To find the mass of \text{H}_2\text{O} needed to produce 25.0 L of \text{O}_2 at STP, we calculated the moles of \text{H}_2\text{O} required (2.232 mol). We then multiplied by the molar mass:
\[2.232 mol \times 18.02 g/mol = 40.2 g\]