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Interpret the equation for the formation of water from its elements in terms of the numbers of molecules, moles, and volumes of gases at STP.

[tex]
2 H_2(g) + O_2(g) \rightarrow 2 H_2O(g)
[/tex]

Sagot :

GinnyAnswer
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.
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