Question

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) can be prepared by heating Si in chlorine gas: $$ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l) $$ In one reaction, \(0.507 \mathrm{~mol}\) of \(\mathrm{SiCl}_{4}\) is produced. How many moles of molecular chlorine were used in the reaction?

Short answer

1.014 moles of \(\mathrm{Cl}_{2}\) were used.

Step-by-step solution

Understand the Stoichiometry of the Reaction

Examine the balanced chemical equation: \[\mathrm{Si}(s) + 2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l)\] This tells us that 1 mole of \(\mathrm{Si}\) reacts with 2 moles of \(\mathrm{Cl}_{2}\) to produce 1 mole of \(\mathrm{SiCl}_{4}\). Thus, the stoichiometric ratio of \(\mathrm{Cl}_{2}\) to \(\mathrm{SiCl}_{4}\) is 2:1.

Set Up the Mole Ratio

The problem gives us that 0.507 moles of \(\mathrm{SiCl}_{4}\) are produced. According to the balanced equation and the stoichiometric ratio, 2 moles of \(\mathrm{Cl}_{2}\) are required to produce 1 mole of \(\mathrm{SiCl}_{4}\).

Calculate Moles of \(\mathrm{Cl}_{2}\) Used

Using the stoichiometric ratio from Step 1, multiply the moles of \(\mathrm{SiCl}_{4}\) by the ratio of \(\mathrm{Cl}_{2}\) to \(\mathrm{SiCl}_{4}\):\[moles\ of\ \mathrm{Cl}_{2} = 0.507\ \text{mol}\ \mathrm{SiCl}_{4}\times \frac{2\ \mathrm{mol}\ \mathrm{Cl}_{2}}{1\ \mathrm{mol}\ \mathrm{SiCl}_{4}} = 1.014\ \mathrm{mol}\ \mathrm{Cl}_{2}\]This calculation shows that 1.014 moles of molecular chlorine were used.

Key concepts

Balanced Chemical Equation

A balanced chemical equation is essential in stoichiometry because it shows the quantitative relationships between reactants and products in a chemical reaction. In the equation \( \mathrm{Si}(s) + 2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l) \), we can see that one mole of silicon (Si) combines with two moles of chlorine gas (\( \mathrm{Cl}_{2} \)) to produce one mole of silicon tetrachloride (\( \mathrm{SiCl}_{4} \)).
This relationship is crucial because it tells us the exact proportions needed for the reaction to proceed without any excess reactants.
To balance an equation, adjust the coefficients (numbers in front of compounds or elements) until the number of atoms of each element is the same on both sides of the equation. This reflects the law of conservation of mass, stating that matter cannot be created or destroyed in a chemical reaction.
Once the equation is balanced, it can be used to calculate how much of a reactant is needed or how much product can be formed during the reaction.

Mole Calculation

Mole calculation is a fundamental part of understanding stoichiometry. The 'mole' is a unit in chemistry that measures the amount of substance, defined as 6.022 x \( 10^{23} \) particles, atoms, or molecules (Avogadro's number).
In our example, we know from the problem statement that 0.507 moles of \( \mathrm{SiCl}_{4} \) are produced. The balanced equation tells us that for every mole of \( \mathrm{SiCl}_{4} \) produced, two moles of \( \mathrm{Cl}_{2} \) are used.
To calculate the moles of \( \mathrm{Cl}_{2} \), utilize the stoichiometric ratio from the balanced equation. Multiply the number of moles of the product by the ratio of \( \mathrm{Cl}_{2} \) to \( \mathrm{SiCl}_{4} \), which is \[2 ext{ moles } \mathrm{Cl}_{2} / 1 ext{ mole } \mathrm{SiCl}_{4}. \] This shows us that \( 0.507 \text{ mol } \mathrm{SiCl}_{4} \times 2 \text{ mol } \mathrm{Cl}_{2} / 1 \text{ mol } \mathrm{SiCl}_{4} = 1.014 \text{ mol } \mathrm{Cl}_{2}. \)
Mole calculations help determine precise amounts of reactants and products and are crucial for any chemical reaction analysis.

Chemical Reaction

A chemical reaction describes the process where substances (reactants) convert into a new form (products). In our reaction, silicon (\( \mathrm{Si} \)) and chlorine gas (\( \mathrm{Cl}_{2} \)) are the reactants, and silicon tetrachloride (\( \mathrm{SiCl}_{4} \)) is the product.
During a chemical reaction, bonds between atoms in the reactants are broken, and new bonds are formed to create the products. The specifics of each reaction are represented in the balanced chemical equation we first discussed.
Chemical reactions can be classified into different types:

• Synthesis Reaction: Two or more simple substances combine to form a more complex product.
• Decomposition Reaction: A single compound breaks down into two or more simpler substances.
• Single Replacement Reaction: One element in a compound is replaced by another element.
• Double Replacement Reaction: The ions of two compounds exchange places to form two new compounds.

In our example, the reaction is a synthesis reaction where \( \mathrm{Si} \) combines with \( \mathrm{Cl}_{2} \) to form \( \mathrm{SiCl}_{4} \). Understanding these processes allows chemists to predict how different chemicals will behave in a reaction and to synthesize new materials.