Questions: Consider the reaction of solid P4 and chlorine gas to form gaseous phosphorus trichloride. Write the balanced chemical equation based on the following description: solid phosphorus reacts with chlorine gas to produce the gaseous phosphorus trichloride. P4(s) + Cl2(g) →

Consider the reaction of solid P4 and chlorine gas to form gaseous phosphorus trichloride.

Write the balanced chemical equation based on the following description: solid phosphorus reacts with chlorine gas to produce the gaseous phosphorus trichloride.
P4(s) + Cl2(g) →

Transcript text: Consider the reaction of solid $\mathrm{P}_{4}$ and chlorine gas to form gaseous phosphorus trichloride. Write the balanced chemical equation based on the following description: solid phosphorus reacts with chlorine gas to produce the gaseous phosphorus trichloride. \[ \mathrm{P}_{4}(\mathrm{~s})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \]

Solution

Solution Steps

Step 1: Identify Reactants and Products

The reactants are solid phosphorus (\(\mathrm{P}_{4}\)) and chlorine gas (\(\mathrm{Cl}_{2}\)). The product is gaseous phosphorus trichloride (\(\mathrm{PCl}_{3}\)).

Step 2: Write the Unbalanced Chemical Equation

The unbalanced chemical equation based on the given reactants and products is: \[ \mathrm{P}_{4}(\mathrm{~s}) + \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g}) \]

Step 3: Balance the Chemical Equation

To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation.

Phosphorus (P): There are 4 phosphorus atoms in \(\mathrm{P}_{4}\) on the reactant side. Therefore, we need 4 phosphorus atoms on the product side, which means we need 4 \(\mathrm{PCl}_{3}\) molecules.
Chlorine (Cl): Each \(\mathrm{PCl}_{3}\) molecule contains 3 chlorine atoms. With 4 \(\mathrm{PCl}_{3}\) molecules, we need \(4 \times 3 = 12\) chlorine atoms. Since chlorine gas (\(\mathrm{Cl}_{2}\)) is diatomic, we need \(\frac{12}{2} = 6\) \(\mathrm{Cl}_{2}\) molecules.

The balanced chemical equation is: \[ \mathrm{P}_{4}(\mathrm{~s}) + 6\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 4\mathrm{PCl}_{3}(\mathrm{~g}) \]

Final Answer

\[ \boxed{\mathrm{P}_{4}(\mathrm{~s}) + 6\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 4\mathrm{PCl}_{3}(\mathrm{~g})} \]

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