Problem 65 - Chemistry and Chemical Reactivity

Question

What volume (in liters) of \(\mathrm{O}_{2}\), measured at standard temperature and pressure, is required to oxidize 0.400 mol of phosphorus \(\left(P_{4}\right) ?\) $$ \mathrm{P}_{4}(\mathrm{s})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{s}) $$

Step-by-Step Solution

Answer

44.8 liters of \(O_{2}\) are needed.

Step 1: Write the Balanced Equation

The balanced chemical equation for the reaction between phosphorus (P_{4}) and oxygen (O_{2}) is given as: \[\mathrm{P}_{4}(\mathrm{s})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{s}) \] This equation indicates that one mole of \(P_{4}\) reacts with five moles of \(O_{2}\) to produce one mole of \(P_{4}O_{10}\).

Step 2: Use Mole Ratios

From the balanced equation, we see that the mole ratio of \(P_{4}\) to \(O_{2}\) is 1:5. Therefore, for every mole of \(P_{4}\) that reacts, 5 moles of \(O_{2}\) are needed. If 0.400 moles of \(P_{4}\) are available, then the moles of \(O_{2}\) required can be calculated as: \[ \text{Moles of } O_{2} = 0.400 \text{ moles of } P_{4} \times \frac{5 \text{ moles of } O_{2}}{1 \text{ mole of } P_{4}} = 2.0 \text{ moles of } O_{2} \]

Step 3: Apply Ideal Gas Law at STP

At standard temperature and pressure (STP), one mole of an ideal gas occupies 22.4 liters. Since we need 2.0 moles of \(O_{2}\), the volume in liters can be calculated using: \[ \text{Volume of } O_{2} = 2.0 \text{ moles} \times 22.4 \text{ liters/mole} = 44.8 \text{ liters} \]

Key Concepts

Understanding Mole Ratios

Mole ratios are a crucial concept in stoichiometry that help us understand the proportions in which substances react or are produced in a chemical reaction. When we have a balanced chemical equation, the coefficients in front of each compound or element tell us the exact mole ratio. In our example, the balanced equation shows that one mole of \( P_4 \) reacts with five moles of \( O_2 \) to form one mole of \( P_4O_{10} \).

• This means if you have 1 mole of phosphorus \( P_4 \), you will need exactly 5 moles of \( O_2 \) to completely react.
• For 0.400 moles of \( P_4 \), we calculate the moles of \( O_2 \) required by multiplying by the ratio \( 5:1 \), giving us 2 moles of \( O_2 \).

Understanding mole ratios helps in predicting how much reactants are needed or products formed in any chemical process. This way, you can plan how much of each substance you'll need in a practical setting.

Balanced Chemical Equations

A balanced chemical equation accurately represents the conservation of mass by showing the same number of each type of atom on both sides of the equation. This is done by adjusting the coefficients (the numbers in front of molecules or atoms) so that the number of atoms for each element is equal on both sides.
In our reaction: \( \mathrm{P}_{4}(\mathrm{s})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{s}) \), the equation is balanced because:

• We have four phosphorus atoms on both the reactant and product sides.
• There are ten oxygen atoms on each side: five \( O_2 \) molecules (totaling ten O atoms) react to form one \( P_4O_{10} \) molecule (which also has ten O atoms).

Balancing equations is essential in stoichiometry as it ensures the law of conservation of mass is not violated. Each element has the same number of atoms before and after the reaction, keeping everything in proper proportion.

The Ideal Gas Law at STP

The Ideal Gas Law is a key principle that relates the pressure, volume, temperature, and amount of a gas. At standard temperature and pressure (STP), which is defined as 0 degrees Celsius (273 K) and 1 atmosphere of pressure, any ideal gas occupies a volume of 22.4 liters per mole.
In our exercise, we used this concept to find out how much volume the required 2 moles of \( O_2 \) would occupy at STP:

• Each mole of \( O_2 \) gas takes up 22.4 liters.
• Therefore, 2 moles would occupy \( 2 \times 22.4 = 44.8 \) liters of volume.

Using the Ideal Gas Law at STP simplifies calculations when dealing with gases in ideal conditions, allowing us to predict volumes and amounts easily. Remembering that 22.4 liters per mole is a good approximation for gas volumes at STP is a handy rule for quickly solving gas stoichiometry problems.