Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our 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.
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.