Question

Elemental phosphorus burns in oxygen with an intensely hot flame, producing a brilliant light and clouds of the oxide product. These properties of the combustion of phosphorus have led to its being used in bombs and incendiary devices for warfare. $$ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{P}_{2} \mathrm{O}_{5}(s) $$ If 4.95 g of phosphorus is burned, what mass of oxygen does it combine with?

Short answer

When 4.95 g of phosphorus is burned, it reacts with 6.40 g of oxygen to form the corresponding oxide product.

Step-by-step solution

Write the balanced chemical equation and identify the given and required information

We are given the balanced chemical equation: \( \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{P}_{2} \mathrm{O}_{5}(s) \) We are also given the mass of phosphorus burned (4.95 g) and asked to find the mass of oxygen consumed during the reaction.

Find the molar masses of P and O

The molar masses of phosphorus (P) and oxygen (O) can be found in the periodic table of elements: - Molar mass of P = 30.97 g/mol - Molar mass of O = 16.00 g/mol

Calculate the moles of P4 from the given mass of P

Using the molar mass of P, calculate the moles of P4 in the given mass of phosphorus: \(\text{moles of P4} = \frac{\text{mass of P}}{\text{molar mass of P4}}\) \(\text{moles of P4} = \frac{4.95\,\text{g}}{4 \times 30.97\,\text{g/mol}} = 0.04\,\text{mol}\)

Convert moles of P4 to moles of O2 using stoichiometry

Use the stoichiometric coefficients from the balanced chemical equation to convert moles of P4 to moles of O2: \(\text{moles of O2} = \text{moles of P4} \times \frac{\text{moles of O2}}{\text{moles of P4}}\) From the balanced equation, we have: \(\frac{5 \text{ moles of O2}}{1 \text{ mole of P4}}\) So, \(\text{moles of O2} = 0.04\,\text{mol} \times \frac{5\,\text{mol O2}}{1\, \text{mol P4}} = 0.20\,\text{mol}\)

Calculate the mass of O2 using its molar mass

Finally, use the molar mass of O2 to find the mass of oxygen consumed: \(\text{mass of O2} = \text{moles of O2} \times \text{molar mass of O2}\) \(\text{mass of O2} = 0.20\,\text{mol} \times 32.00\,\text{g/mol} = 6.40\,\text{g}\) So, 6.40 grams of oxygen are combined with 4.95 grams of phosphorus during the combustion reaction.

Key concepts

Chemical Reactions

Chemical reactions involve rearrangement of atoms to transform reactants into products. This process involves breaking bonds in reactants and forming new ones in products. In the given exercise, elemental phosphorus (\( \mathrm{P}_4 \)) reacts with oxygen (\( \mathrm{O}_2 \)) to form diphosphorus pentoxide (\( \mathrm{P}_2\mathrm{O}_5 \)).
This transformation is a combustion reaction, characterized by phosphorus burning with oxygen to produce a new compound.

• The reaction releases energy in the form of heat and light.
• It shows the interaction between a non-metal (phosphorus) and a gas (oxygen).

Understanding chemical reactions requires knowing:

• The reactants that combine in the reaction, like phosphorus and oxygen in the example.
• The products formed, which is \( \mathrm{P}_2\mathrm{O}_5 \) in this case.
• The type of reaction, e.g., combustion.

Molar Mass

Molar mass is the mass of one mole of a substance and is crucial for converting between grams and moles. It is determined using the periodic table where the atomic mass of each element is found.
For example:

• The molar mass of phosphorus (P) is 30.97 g/mol.
• The molar mass of oxygen (O) is 16.00 g/mol.

When calculating molar masses for compounds:

• Multiply the molar mass of each element by the number of atoms of that element in a molecule.
• Add these values together.

For \( \mathrm{O}_2 \), the calculation is straightforward:\[ \text{Molar mass of } \mathrm{O}_2 = 2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol} \] This concept allows us to convert between mass in grams and moles, which is essential for stoichiometry and correctly quantifying chemical reactions.

Balancing Chemical Equations

Balancing chemical equations ensures the conservation of mass where the number of atoms for each element remains the same on both sides of the equation. For the reaction given in the exercise:\[ \mathrm{P}_{4}(s) + 5 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{P}_{2} \mathrm{O}_{5}(s) \]We identify that:

• 4 phosphorus atoms are in both reactants and products.
• There are 10 oxygen atoms on each side as well.

Balancing involves:

• Counting the atoms of each element in the reactants and products.
• Adjusting the coefficients to ensure that each side has the same number of atoms.

The coefficients (like 5 for \( \mathrm{O}_2 \)) are crucial as they dictate how many molecules of each substance take part in the reaction, ultimately informing us about the stoichiometry of the reaction, or the quantitative relationships between reactants and products.