Question

5.53 How many grams of phosphorus react with \(35.5 \mathrm{~L}\) of \(\mathrm{O}_{2}\) at STP to form tetraphosphorus decoxide? $$\mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)$$

Short answer

39.29 g of phosphorus.

Step-by-step solution

- Determine moles of oxygen gas

Use the fact that at STP, 1 mole of any gas occupies 22.4 L. Calculate the moles of \(\text{O}_2\) using the given volume. \[ \text{Moles of } \text{O}_2 = \frac{35.5 \text{ L}}{22.4 \text{ L/mol}} = 1.584 \text{ mol} \]

- Use the balanced equation

From the balanced chemical equation, \(\text{P}_4 + 5 \text{O}_2 \rightarrow \text{P}_4\text{O}_{10}\), 5 moles of \(\text{O}_2\) react with 1 mole of \(\text{P}_4\). Set up the ratio to find the moles of \(\text{P}_4\). \[ \text{Moles of } \text{P}_4 = \frac{1.584 \text{ mol } \text{O}_2}{5} = 0.317 \text{ mol } \text{P}_4 \]

- Find the mass of phosphorus

Multiply the moles of \(\text{P}_4\) by the molar mass of \(\text{P}_4\) (123.9 g/mol). \[ \text{Mass of } \text{P}_4 = 0.317 \text{ mol } \text{P}_4 \times 123.9 \text{ g/mol} = 39.29 \text{ g} \]

Key concepts

Moles and Molar Mass

In chemistry, understanding moles and molar mass is essential. A mole is a unit that measures the amount of a substance. One mole is equivalent to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities (such as atoms, molecules, or ions).
For example, one mole of carbon atoms equals \(6.022 \times 10^{23}\) carbon atoms.
The molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol).
It's calculated by adding up the atomic masses of all the atoms in a molecule.
If you know the molar mass of a substance and the amount in moles, you can easily find the total mass with the formula:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]
In our exercise, the problem asks for the mass of phosphorus. Knowing the molar mass of \( \text{P}_4 \) is 123.9 g/mol allows us to convert moles of \( \text{P}_4 \) to grams.

Chemical Reaction Equation

A chemical reaction equation symbolizes the transformation of reactants into products.
It must be balanced to comply with the law of conservation of mass, meaning the number of atoms of each element should be the same on both sides. A balanced equation tells us the ratios of reactants to products.
For example, in our exercise, the balanced equation is:
\[ \text{P}_4(s) + 5 \text{O}_2(g) \rightarrow \text{P}_4\text{O}_{10}(s) \]
This equation indicates that 1 mole of tetraphosphorus reacts with 5 moles of oxygen gas to produce 1 mole of tetraphosphorus decoxide.
This ratio is crucial for stoichiometric calculations as it allows us to convert between moles of different substances.

STP Conditions

STP stands for Standard Temperature and Pressure.
It is a common reference point in chemistry for calculations involving gases.
At STP, the temperature is 0°C (273.15 K) and the pressure is 1 atm.
Under these conditions, one mole of any ideal gas occupies 22.4 liters.
This conversion factor simplifies calculations involving gases, as seen in our exercise. To find the moles of oxygen gas from the volume given, we use the relationship:
\[ \text{Moles of } \text{O}_2 = \frac{35.5 \text{ L}}{22.4 \text{ L/mol}} \]
Understanding STP conditions helps in converting gas volumes to moles, a critical step in many stoichiometry problems.