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Chapter 8: Problem 142

When you strike a match, trace amounts of white phosphorus \(\left(\mathrm{P}_{4}\right)\) are converted into an oxide of phosphorus, \(\mathrm{P}_{4} \mathrm{O}_{10}\). This conversion releases sufficient heat to ignite the match head. Write a balanced reaction for this conversion, and classify the reaction type(s).

Short Answer

Expert verified
The balanced equation for the conversion of white phosphorus (P4) into an oxide of phosphorus (P4O10) is \( \mathrm{P}_4 + 5\mathrm{O}_2 \rightarrow \mathrm{P}_4\mathrm{O}_{10} \). The reaction type is Synthesis (Combination).

Step by step solution

01

Write down the unbalanced chemical equation

Firstly, write the unbalanced chemical equation for the given conversion: \[ \mathrm{P}_4 + \mathrm{O}_2 \rightarrow \mathrm{P}_4\mathrm{O}_{10} \]
02

Balance the chemical equation

In order to balance the chemical equation, we need to make sure that the number of atoms for each element is the same on both sides of the equation. We can see that there are 10 oxygen atoms in P4O10 on the right side of the equation, so we need 5 molecules of O2 to balance the oxygen atoms: \[ \mathrm{P}_4 + 5\mathrm{O}_2 \rightarrow \mathrm{P}_4\mathrm{O}_{10} \] Now the equation is balanced because there are 4 atoms of phosphorus and 10 atoms of oxygen on both sides.
03

Classify the reaction type(s)

This reaction is a combination of two elements, phosphorus and oxygen, to form a compound, P4O10. Therefore, this is a synthesis (combination) reaction. Combining the steps, we can conclude: Balanced equation: \( \mathrm{P}_4 + 5\mathrm{O}_2 \rightarrow \mathrm{P}_4\mathrm{O}_{10} \) Reaction type: Synthesis (Combination)

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