At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
Absolutely, let's delve into the given balanced chemical equation for the formation of water from hydrogen and oxygen:
[tex]\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g) \][/tex]
We'll interpret this equation in terms of numbers of molecules, moles, and volumes of gases at Standard Temperature and Pressure (STP).
### Numbers of Molecules
First, consider the number of individual molecules:
1. Reactants:
- [tex]\(2\)[/tex] molecules of [tex]\(H_2\)[/tex]
- [tex]\(1\)[/tex] molecule of [tex]\(O_2\)[/tex]
2. Products:
- [tex]\(2\)[/tex] molecules of [tex]\(H_2O\)[/tex]
So, the reaction indicates that 2 molecules of hydrogen gas ([tex]\(H_2\)[/tex]) react with 1 molecule of oxygen gas ([tex]\(O_2\)[/tex]) to form 2 molecules of water ([tex]\(H_2O\)[/tex]).
### Moles
Next, let's talk about moles. The coefficients in a balanced chemical equation also represent the ratios of the substances in moles:
1. Reactants:
- [tex]\(2\)[/tex] moles of [tex]\(H_2\)[/tex]
- [tex]\(1\)[/tex] mole of [tex]\(O_2\)[/tex]
2. Products:
- [tex]\(2\)[/tex] moles of [tex]\(H_2O\)[/tex]
Thus, in this reaction, 2 moles of hydrogen gas ([tex]\(H_2\)[/tex]) react with 1 mole of oxygen gas ([tex]\(O_2\)[/tex]) to produce 2 moles of water ([tex]\(H_2O\)[/tex]).
### Volumes of Gases at STP
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies [tex]\(22.4\)[/tex] liters. Thus, we can calculate the volume of each gas involved in the reaction:
1. Reactants:
- Volume of [tex]\(H_2\)[/tex]: [tex]\(2 \text{ moles} \times 22.4 \text{ liters/mole} = 44.8\)[/tex] liters
- Volume of [tex]\(O_2\)[/tex]: [tex]\(1 \text{ mole} \times 22.4 \text{ liters/mole} = 22.4\)[/tex] liters
2. Products:
- Volume of [tex]\(H_2O\)[/tex]: [tex]\(2 \text{ moles} \times 22.4 \text{ liters/mole} = 44.8\)[/tex] liters
Therefore, according to the equation, 44.8 liters of hydrogen gas ([tex]\(H_2\)[/tex]) react with 22.4 liters of oxygen gas ([tex]\(O_2\)[/tex]) to produce 44.8 liters of water vapor ([tex]\(H_2O\)[/tex]) at STP.
This interpretation comprehensively covers the molecule counts, moles, and gas volumes at standard conditions, aligning with the balanced chemical equation provided.
[tex]\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g) \][/tex]
We'll interpret this equation in terms of numbers of molecules, moles, and volumes of gases at Standard Temperature and Pressure (STP).
### Numbers of Molecules
First, consider the number of individual molecules:
1. Reactants:
- [tex]\(2\)[/tex] molecules of [tex]\(H_2\)[/tex]
- [tex]\(1\)[/tex] molecule of [tex]\(O_2\)[/tex]
2. Products:
- [tex]\(2\)[/tex] molecules of [tex]\(H_2O\)[/tex]
So, the reaction indicates that 2 molecules of hydrogen gas ([tex]\(H_2\)[/tex]) react with 1 molecule of oxygen gas ([tex]\(O_2\)[/tex]) to form 2 molecules of water ([tex]\(H_2O\)[/tex]).
### Moles
Next, let's talk about moles. The coefficients in a balanced chemical equation also represent the ratios of the substances in moles:
1. Reactants:
- [tex]\(2\)[/tex] moles of [tex]\(H_2\)[/tex]
- [tex]\(1\)[/tex] mole of [tex]\(O_2\)[/tex]
2. Products:
- [tex]\(2\)[/tex] moles of [tex]\(H_2O\)[/tex]
Thus, in this reaction, 2 moles of hydrogen gas ([tex]\(H_2\)[/tex]) react with 1 mole of oxygen gas ([tex]\(O_2\)[/tex]) to produce 2 moles of water ([tex]\(H_2O\)[/tex]).
### Volumes of Gases at STP
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies [tex]\(22.4\)[/tex] liters. Thus, we can calculate the volume of each gas involved in the reaction:
1. Reactants:
- Volume of [tex]\(H_2\)[/tex]: [tex]\(2 \text{ moles} \times 22.4 \text{ liters/mole} = 44.8\)[/tex] liters
- Volume of [tex]\(O_2\)[/tex]: [tex]\(1 \text{ mole} \times 22.4 \text{ liters/mole} = 22.4\)[/tex] liters
2. Products:
- Volume of [tex]\(H_2O\)[/tex]: [tex]\(2 \text{ moles} \times 22.4 \text{ liters/mole} = 44.8\)[/tex] liters
Therefore, according to the equation, 44.8 liters of hydrogen gas ([tex]\(H_2\)[/tex]) react with 22.4 liters of oxygen gas ([tex]\(O_2\)[/tex]) to produce 44.8 liters of water vapor ([tex]\(H_2O\)[/tex]) at STP.
This interpretation comprehensively covers the molecule counts, moles, and gas volumes at standard conditions, aligning with the balanced chemical equation provided.
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.