Problem 16 - General Chemistry: Principles and Modern Applications

Question

If \(46.3 \mathrm{g} \mathrm{PCl}_{3}\) is produced by the reaction $$ 6 \mathrm{Cl}_{2}(\mathrm{g})+\mathrm{P}_{4}(\mathrm{s}) \longrightarrow 4 \mathrm{PCl}_{3}(\mathrm{l}) $$ how many grams each of \(\mathrm{Cl}_{2}\) and \(\mathrm{P}_{4}\) are consumed?

Step-by-Step Solution

Answer

The consumed mass of \(Cl_{2}\) and \(P_{4}\) are 35.84 g and 10.43 g respectively.

Step 1: Calculate Moles of \(PCl_{3}\)

Firstly, calculate the number of moles of \(PCl_{3}\) produced using the given weight and the molar mass of \(PCl_{3}\). The molar mass of \(PCl_{3}\) is approximately \(30.97+35.45*3 = 137.32 \, g/mol\). So, number of moles = mass/molar mass = \(46.3/137.32 = 0.337 \, mol\)

Step 2: Determine Moles of \(Cl_{2}\) and \(P_{4}\)

Next, use the stoichiometry of the chemical reaction to calculate the moles of \(Cl_{2}\) and \(P_{4}\). From the balanced equation, we can see for every 4 moles of \(PCl_{3}\) produced, 6 moles of \(Cl_{2}\) are consumed and 1 mole of \(P_{4}\) is consumed. Therefore, \(Cl_{2} = 6/4 \times 0.337 = 0.5055 \, mol\) and \(P_{4} = 1/4 \times 0.337 = 0.08425 \, mol\)

Step 3: Calculate Mass of Consumed Reactants

We then convert these mole quantities back to mass using the molar mass of each substance. The molar mass of \(Cl_{2}\) is \(35.45*2 = 70.9 \, g/mol\), and that of \(P_{4}\) is \(30.97*4 = 123.88 \, g/mol\). Therefore, mass of \(Cl_{2} = 0.5055 \times 70.9 = 35.84 \, g\) and mass of \(P_{4} = 0.08425 \times 123.88 = 10.43 \, g\)

Key Concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into new substances known as products. In these reactions, chemical bonds are broken and formed, ultimately changing the molecular structure of the substances involved. Every chemical reaction can be represented by a balanced chemical equation that shows the reactants and products.
In the given example, **chlorine gas** (\(Cl_2\)) reacts with **phosphorus** (\(P_4\)) to form **phosphorus trichloride** (\(PCl_3\)). This reaction can be represented by the balanced equation:
\[ 6\, Cl_2(g) + P_4 (s) \rightarrow 4\, PCl_3(l) \]
The balanced equation ensures the conservation of mass, meaning the number of atoms of each element is the same on both sides of the equation. This principle is vital for performing stoichiometric calculations, guiding us on how much of each reactant is needed to produce a desired amount of product.

Molar Mass

Molar mass is an essential concept in chemistry and is used to convert between the mass of a substance and the number of moles. It is the mass of one mole of a given substance, typically measured in grams per mole (g/mol). Molar mass allows chemists to measure out substances in a laboratory setting accurately.
For example, the molar mass of \(PCl_3\), as shown in our exercise, is calculated as follows:

• The atomic mass of phosphorus (P) is approximately 30.97 g/mol.
• The atomic mass of chlorine (Cl) is around 35.45 g/mol.

Hence, the molar mass of \(PCl_3\) is \(30.97 + 3\times35.45 = 137.32\, g/mol\). This calculation is crucial in determining how much of a reactant forms a specific amount of product.
Similarly, for \(Cl_2\), the molar mass is \(2\times35.45 = 70.9\, g/mol\), and for \(P_4\), it is \(4\times30.97 = 123.88\, g/mol\). Knowing these values allows conversion of moles to grams, facilitating precise measurements in chemical reactions.

Mole Calculations

Mole calculations allow chemists to quantify the amount of substance involved in a reaction. The concept of the mole is fundamental in chemistry, where 1 mole is equivalent to \(6.022 \times 10^{23}\) entities (Avogadro's number). When dealing with reactions, knowing the number of moles helps determine the proportions of reactants and products involved.
In the exercise, we start by calculating the moles of \(PCl_3\) produced:
\[ \text{Moles of } PCl_3 = \frac{46.3\, g}{137.32\, g/mol} = 0.337\, mol\]
Next, using the stoichiometry from the balanced equation, we calculate the moles of \(Cl_2\) and \(P_4\) consumed:

• From the balanced equation, every 4 moles of \(PCl_3\) needs 6 moles of \(Cl_2\). Thus, moles of \(Cl_2\):
• \(Cl_2 = \frac{6}{4} \times 0.337 = 0.5055\, mol\)
• Similarly, every 4 moles of \(PCl_3\) corresponds to 1 mole of \(P_4\). Hence, moles of \(P_4\):
• \(P_4 = \frac{1}{4} \times 0.337 = 0.08425\, mol\)

Converting these moles back to grams, using their respective molar masses, gives the amounts of reactants consumed in the reaction. Understanding these calculations is critical to accurately determining how much raw material is needed to achieve a desired amount of product.