Problem 77 - Chemistry The Molecular Science

Question

The ceramic silicon nitride, \(\mathrm{Si}_{3} \mathrm{~N}_{4}\), is made by heating silicon and nitrogen at an elevated temperature. $$3 \mathrm{Si}(\mathrm{s})+2 \mathrm{~N}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Si}_{3} \mathrm{~N}_{4}(\mathrm{~s})$$ Calculate the mass of silicon that must combine with excess \(\mathrm{N}_{2}\) to produce \(1.0 \mathrm{~kg} \mathrm{Si}_{3} \mathrm{~N}_{4}\) if this process is \(92 \%\) efficient.

Step-by-Step Solution

Answer

652.09 g of silicon is needed.

Step 1: Calculate the molar mass of Si₃N₄

First, determine the molar mass of silicon nitride, \( \mathrm{Si}_3\mathrm{N}_4 \). Silicon (\( \mathrm{Si} \)) has a molar mass of approximately 28.09 \( \mathrm{g/mol} \) and nitrogen (\( \mathrm{N} \)) has a molar mass of approximately 14.01 \( \mathrm{g/mol} \). Thus, the molar mass of \( \mathrm{Si}_3\mathrm{N}_4 \) is calculated as follows:\[3 \times 28.09 + 4 \times 14.01 = 140.28 \, \mathrm{g/mol}\]

Step 2: Calculate moles of Si₃N₄ needed

We need to find out how many moles of \( \mathrm{Si}_3\mathrm{N}_4 \) correspond to 1.0 kg. Convert 1.0 kg into grams, which is 1000 g, and then calculate the moles:\[\text{Moles of } \mathrm{Si}_3\mathrm{N}_4 = \frac{1000 \, \mathrm{g}}{140.28 \, \mathrm{g/mol}} = 7.13 \, \mathrm{mol}\]

Step 3: Account for the efficiency of the reaction

Since the reaction is 92% efficient, we need to adjust the moles of \( \mathrm{Si}_3\mathrm{N}_4 \) by dividing by the efficiency:\[\text{Effective moles of } \mathrm{Si}_3\mathrm{N}_4 = \frac{7.13 \, \mathrm{mol}}{0.92} = 7.75 \, \mathrm{mol}\]

Step 4: Determine moles of Si required

From the balanced chemical equation:\( 3 \mathrm{Si} + 2 \mathrm{~N}_2 \rightarrow \mathrm{Si}_3\mathrm{N}_4 \), 3 moles of \( \mathrm{Si} \) are needed for every 1 mole of \( \mathrm{Si}_3\mathrm{N}_4 \). Therefore, calculate the moles of \( \mathrm{Si} \) required:\[\text{Moles of Si required} = 3 \times 7.75 \, \mathrm{mol} = 23.25 \, \mathrm{mol}\]

Step 5: Calculate the mass of Si needed

Finally, calculate the mass of \( \mathrm{Si} \) using its molar mass (28.09 \( \mathrm{g/mol} \)):\[\text{Mass of Si} = 23.25 \, \mathrm{mol} \times 28.09 \, \mathrm{g/mol} = 652.09 \, \mathrm{g}\]

Key Concepts

Chemical Reactions

Chemical reactions describe the transformation of substances through the breaking and forming of bonds. In many reactions, such as the one given here, a set amount of reactants combines to form products. Understanding the stoichiometry of a reaction, like that of silicon (\( \text{Si} \)) and nitrogen (\( \text{N}_2 \)) forming silicon nitride (\( \text{Si}_3\text{N}_4 \)), is important to predict how much of each substance is needed.

In the balanced chemical equation provided, \[ 3 \text{Si} + 2 \text{N}_2 \rightarrow \text{Si}_3\text{N}_4 \] we see that 3 moles of silicon react with 2 moles of nitrogen gas to form 1 mole of silicon nitride.

• The reactants participate in a fixed ratio as dictated by the equation.
• This ratio helps determine the amount of each reactant needed for a certain quantity of product.

The stoichiometric coefficients (the numbers in front of \( \text{Si} \) and \( \text{N}_2 \)) help explain the quantitative relationships in this reaction. Skewing these proportions would mean incomplete reactions or excess of one reactant, leading to inefficiencies.

Molar Mass

Molar mass links the number of moles of a substance to its mass in grams. This is crucial because chemical equations work in moles, but laboratory measurements are usually in grams. Let's see how you can use molar mass to transition between these units.

For silicon nitride (\( \text{Si}_3\text{N}_4 \)), the molar mass is calculated by summing the molar masses of silicon and nitrogen:

• Silicon (\( \text{Si} \)) has a molar mass of 28.09 g/mol.
• Nitrogen (\( \text{N} \)) has a molar mass of 14.01 g/mol.
• Thus, \( \text{Si}_3\text{N}_4 \) is calculated as: \[3 \times 28.09 + 4 \times 14.01 = 140.28 \, \text{g/mol}\]

This calculated molar mass acts as a bridge to convert between moles of \( \text{Si}_3\text{N}_4 \) and mass: by dividing or multiplying, you switch between g and mol, aiding in understanding how much reactant is necessary or product is produced.

Reaction Efficiency

Reaction efficiency is a practical tool to measure how effectively reactants are converted into products. It considers factors like impurities or incomplete reactions that prevent 100% yield.

In this exercise, we see a reaction efficiency of 92%, which reflects that only 92% of potential yield is realistic. To account for this, calculated moles or mass should be adjusted:

• Using the formula: \[ \text{Actual Yield} = \text{Theoretical Yield} \times \text{Efficiency} \]
• Given \(7.13\, \text{mol} \) as theoretical, dividing by 0.92 results in \(7.75\, \text{mol} \) as necessary for achieving the desired output.

Reaction efficiency is crucial in industrial applications where reactants cost money and waste minimization is desired. For students, understanding the impact of efficiency in chem equations fosters a deeper appreciation for real-world chemical processes.