Problem 32 - Chemistry and Chemical Reactivity

Question

Silane, \(\operatorname{SiH}_{4},\) reacts with \(\mathrm{O}_{2}\) to give silicon dioxide and water: $$\mathrm{SiH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SiO}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$$ A \(5.20-\) L sample of \(\mathrm{SiH}_{4}\) gas at \(356 \mathrm{mm}\) Hg pressure and \(25^{\circ} \mathrm{C}\) is allowed to react with \(\mathrm{O}_{2}\) gas. What volume of O, gas, in liters, is required for complete reaction if the oxygen has a pressure of \(425 \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

8.68 liters of \( \mathrm{O}_2 \) are required.

Step 1: Use the Ideal Gas Law to Find Moles of SiH4

First, we convert the pressure from mm Hg to atm: \( P = \frac{356 \text{ mm Hg}}{760 \text{ mm Hg/atm}} = 0.468 \text{ atm} \). Temperature in Kelvin: \( T = 25^{\circ}C + 273 = 298 K \). Use the ideal gas law, \( PV = nRT \), to find the moles of \( \mathrm{SiH}_4 \): \[ n = \frac{PV}{RT} = \frac{(0.468 \text{ atm})(5.20 \text{ L})}{0.0821 \text{ L atm/mol K} \times 298 \text{ K}} = 0.099 \text{ moles}. \]

Step 2: Use Stoichiometry to Find Moles of O2 Required

According to the balanced equation, 1 mole of \( \mathrm{SiH}_4 \) reacts with 2 moles of \( \mathrm{O}_2 \). Calculate moles of \( \mathrm{O}_2 \) needed: \[ \text{moles of } \mathrm{O}_2 = 2 \times 0.099 = 0.198 \text{ moles}. \]

Step 3: Use the Ideal Gas Law to Find Volume of O2 Required

Given \( P = \frac{425 \text{ mm Hg}}{760 \text{ mm Hg/atm}} = 0.559 \text{ atm} \) and \( T = 298 K \). Rearrange the ideal gas law to find volume: \[ V = \frac{nRT}{P} = \frac{(0.198 \text{ moles})(0.0821 \text{ L atm/mol K})(298 \text{ K})}{0.559 \text{ atm}} = 8.68 \text{ L}. \]

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves calculating the relationships between reactants and products in a chemical reaction. It is like the recipe in cooking, ensuring all ingredients are in the correct proportion.

In our reaction between silane \(\mathrm{SiH}_4\) and \(\mathrm{O}_2\), we use stoichiometry to understand how much of each reactant is needed and what the products will be. According to the balanced chemical equation, 1 mole of silicon hydride reacts with 2 moles of oxygen gas to produce silicon dioxide and water:

\( \mathrm{SiH_4} + 2 \mathrm{O_2} \rightarrow \mathrm{SiO_2} + 2 \mathrm{H_2O} \)

This tells us that for every mole of \(\mathrm{SiH}_4\), 2 moles of \(\mathrm{O}_2\) must react. Stoichiometry lets us calculate how much \(\mathrm{O}_2\) is required if we have a specific amount of \(\mathrm{SiH}_4\), ensuring we have enough material to fully complete the reaction without any leftovers or shortages.

This is crucial in reactions because it guides the efficient use of chemicals, minimizing waste and cost. And remember, balancing chemical equations is always a prerequisite to performing stoichiometric calculations!

Gas Pressure Conversion

Gas pressure conversion is an important task in calculations involving gases, especially when using the ideal gas law.

Pressure can be measured in various units, such as millimeters of mercury (mm Hg), atmospheres (atm), or pascals (Pa). Since calculations often require specific units, converting between these is necessary. In this problem, we start with \(356 \, \mathrm{mm} \space \mathrm{Hg}\) for \(\mathrm{SiH}_4\) and \(425 \, \mathrm{mm} \space \mathrm{Hg}\) for \(\mathrm{O}_2\).

To convert mm Hg to atm, use the conversion factor:

• 1 atmosphere = 760 mm Hg.

Applying this, the pressure of \(\mathrm{SiH}_4\) transforms to \(0.468 \, \text{atm}\), while \(\mathrm{O}_2\) transforms to \(0.559 \, \text{atm}\). With these conversions, they conveniently fit into the ideal gas law equation.

Such conversions are vital because they allow consistent and accurate use of the gas laws in calculations across different conditions.

Balanced Chemical Equation

A balanced chemical equation is a cornerstone in chemistry that accurately represents the conservation of mass in reactions. It ensures that the number of each type of atom is the same on both sides of the equation.

For the reaction of silane with oxygen:
\[ \mathrm{SiH_4} + 2 \mathrm{O_2} \rightarrow \mathrm{SiO_2} + 2 \mathrm{H_2O} \]

This equation is balanced because it has equal numbers of silicon, hydrogen, and oxygen atoms on both the reactant and product sides.

Balancing is often the first step before performing any calculations. It allows us to use stoichiometry effectively, establishing mole ratios between the different substances.

In our exercise, this ratio tells us directly how many moles of \(\mathrm{O}_2\) we will need to completely react with the given moles of \(\mathrm{SiH}_4\). Without a balanced equation, any stoichiometric calculation would be incomplete and likely inaccurate. So, remember always to check and balance your equation!

Gas Volume Calculation

Gas volume calculation often involves using the ideal gas law when conditions like temperature and pressure are known.

The ideal gas law, given by:
\[ PV = nRT \]

relates pressure (P), volume (V), number of moles (n), the ideal gas constant (R), and temperature (T).

In this task, once we know the number of moles of \(\mathrm{O}_2\) needed (0.198 moles from stoichiometric calculations) and the pressure of \(\mathrm{O}_2\) (0.559 atm), we rearrange the equation to solve for the volume of \(\mathrm{O}_2\):
\[ V = \frac{nRT}{P} \]

Using:

• \(n = 0.198 \text{ moles} \)
• \(R = 0.0821 \text{ L atm/mol K} \)
• \(T = 298 \text{ K} \)
• \(P = 0.559 \text{ atm} \)

We find the volume of \(\mathrm{O}_2\) required is \(8.68 \text{ L}\).

This fortifies the importance of understanding how we can manipulate the ideal gas law to find various missing pieces in a chemical systems puzzle, like volume, pressure, or the number of moles.