Problem 41
Question
Gaseous silane, \(\mathrm{SiH}_{4}\), ignites spontaneously in air according to the equation $$\mathrm{SiH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SiO}_{2}(\mathrm{~s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ If \(5.2 \mathrm{~L} \mathrm{SiH}_{4}\) reacts with \(\mathrm{O}_{2},\) determine the volume in liters of \(\mathrm{O}_{2}\) required for complete reaction. Determine the volume of \(\mathrm{H}_{2} \mathrm{O}\) vapor produced. Assume all gases are measured at the same temperature and pressure.
Step-by-Step Solution
Verified Answer
10.4 L of \( \mathrm{O}_{2} \) is required and 10.4 L of \( \mathrm{H}_{2} \mathrm{O} \) vapor is produced.
1Step 1: Understand the reaction stoichiometry
The balanced chemical equation is \( \mathrm{SiH}_{4}(\mathrm{~g}) + 2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SiO}_{2}(\mathrm{~s}) + 2 \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \). This tells us that 1 mole (or liter, since gases at the same temperature and pressure can be compared by volume) of \( \mathrm{SiH}_{4} \) reacts with 2 moles (or liters) of \( \mathrm{O}_{2} \) to produce 1 mole (solid) of \( \mathrm{SiO}_{2} \) and 2 moles (or liters) of \( \mathrm{H}_{2}\mathrm{O} \) vapor.
2Step 2: Calculate the volume of O2 required
Given \( 5.2 \mathrm{~L} \) of \( \mathrm{SiH}_{4} \) reaction with \( \mathrm{O}_{2} \), use the stoichiometry: For every 1 L of \( \mathrm{SiH}_{4} \), 2 L of \( \mathrm{O}_{2} \) are required. Thus, \( 5.2 \mathrm{~L} \) of \( \mathrm{SiH}_{4} \) will require \( 5.2 \mathrm{~L} \times 2 = 10.4 \mathrm{~L} \) of \( \mathrm{O}_{2} \).
3Step 3: Calculate the volume of H2O vapor produced
According to stoichiometry, 1 L of \( \mathrm{SiH}_{4} \) produces 2 L of \( \mathrm{H}_{2}\mathrm{O} \) vapor. Thus, from \( 5.2 \mathrm{~L} \) of \( \mathrm{SiH}_{4} \), the volume of \( \mathrm{H}_{2}\mathrm{O} \) vapor produced is \( 5.2 \mathrm{~L} \times 2 = 10.4 \mathrm{~L} \).
Key Concepts
Gas ReactionsChemical EquationsVolume Calculations
Gas Reactions
Gas reactions describe chemical processes that occur between gaseous substances. In such reactions, molecules of gases interact to form new products. Understanding these reactions is crucial because they often involve changes in the volumes of reactants and products under specific conditions of temperature and pressure.
Consider the reaction of silane (\(\mathrm{SiH}_{4}\)) with oxygen (\(\mathrm{O}_{2}\)) as an example. This reaction results in the formation of silicon dioxide (\(\mathrm{SiO}_{2}\)) and water vapor (\(\mathrm{H}_{2}\mathrm{O}\)). The balanced chemical equation represents this gas reaction:
It's interesting to note that in gas reactions, conditions such as ideal gas behavior, temperature, and pressure are assumed to facilitate easier calculations. Once you grasp these fundamental ideas, finding the volume of gases needed or produced becomes much simpler.
Consider the reaction of silane (\(\mathrm{SiH}_{4}\)) with oxygen (\(\mathrm{O}_{2}\)) as an example. This reaction results in the formation of silicon dioxide (\(\mathrm{SiO}_{2}\)) and water vapor (\(\mathrm{H}_{2}\mathrm{O}\)). The balanced chemical equation represents this gas reaction:
- \(\mathrm{SiH}_{4}(\mathrm{~g}) + 2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SiO}_{2}(\mathrm{~s}) + 2 \mathrm{H}_{2}\mathrm{O}(\mathrm{g})\)
It's interesting to note that in gas reactions, conditions such as ideal gas behavior, temperature, and pressure are assumed to facilitate easier calculations. Once you grasp these fundamental ideas, finding the volume of gases needed or produced becomes much simpler.
Chemical Equations
A chemical equation is a symbolic representation of a chemical reaction. It shows which substances react together and what products they form. Understanding chemical equations is essential because they tell us about the conservation of mass and the stoichiometric relationships between reactants and products.
In the case of our silane reaction, the equation:
These coefficients are vital for stoichiometric calculations, allowing us to determine the exact amount of each substance involved. In essence, balancing chemical equations helps us understand the conservation of atoms, ensuring no atoms are lost or gained during the reaction.
In the case of our silane reaction, the equation:
- \(\mathrm{SiH}_{4}(\mathrm{~g}) + 2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SiO}_{2}(\mathrm{~s}) + 2 \mathrm{H}_{2}\mathrm{O}(\mathrm{g})\)
- "Reactants" like \(\mathrm{SiH}_{4}\) and \(\mathrm{O}_{2}\) are on the left side.
- "Products" such as \(\mathrm{SiO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) are on the right side.
- The numbers before the formulas, known as coefficients, indicate how many molecules participate in the reaction.
These coefficients are vital for stoichiometric calculations, allowing us to determine the exact amount of each substance involved. In essence, balancing chemical equations helps us understand the conservation of atoms, ensuring no atoms are lost or gained during the reaction.
Volume Calculations
Volume calculations in stoichiometry help us understand how much gas is involved in reactant or product form during a chemical reaction. This aspect of chemistry allows us to harness the predictability of reactions happening under controlled conditions.
For reactions involving gases, it’s convenient to use the concept of molar volume. Under the same conditions of temperature and pressure, the volumes of gaseous reactants and products will follow the mole ratios given by the stoichiometry of the chemical equation.
In the reaction of silane with oxygen, the volume calculations would follow the balanced equation:
For reactions involving gases, it’s convenient to use the concept of molar volume. Under the same conditions of temperature and pressure, the volumes of gaseous reactants and products will follow the mole ratios given by the stoichiometry of the chemical equation.
In the reaction of silane with oxygen, the volume calculations would follow the balanced equation:
- \(\mathrm{SiH}_{4}(\mathrm{~g}) + 2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SiO}_{2}(\mathrm{~s}) + 2 \mathrm{H}_{2}\mathrm{O}(\mathrm{g})\)
- 5.2 liters of \(\mathrm{SiH}_{4}\) will need 2 times that volume of \(\mathrm{O}_{2}\) (10.4 liters) to react completely.
- Similarly, 5.2 liters of \(\mathrm{SiH}_{4}\) will produce 10.4 liters of \(\mathrm{H}_{2}\mathrm{O}\) vapor.
Other exercises in this chapter
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