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MatroZZZ [7]
3 years ago
6

Why in a stream containing water, the mole fraction of a given

Chemistry
1 answer:
Rashid [163]3 years ago
6 0

Answer:

A liquid, at any temperature, is in equilibrium with its own steam. This means that on the surface of the liquid or solid substance, there are gaseous molecules of this substance. These molecules exert a pressure on the liquid phase, a pressure known as vapor pressure.

In chemistry, when we talk about dry basis, we talk about a state in which the presence of water in a gaseous state is denied for the calculation. So vapor pressure equals zero.

When we talk about the wet basis, the presence of water in the steam is considered for the calculation, which normally is expressed as a percentage or moisture.

In summary, for a gas mixture steam:

  • For dry basis, we just have <em>component A, component B....</em>
  • For wet basis, we have <em>water vapor, component A, component B...</em>

So, in wet basis we have an extra component (water).

Assuming we only have 2 components in our steam, and being X the molar fraction of eact component:

  • For dry basis: Xa + Xb = 1................................. Xa = 1 - Xb
  • For wet basis: Xa + Xb + Xwater = 1 .............Xa = 1 - Xwater - Xb

For dry basis the mole fraction of A it is obtained by subtracting the molar fraction of B from one. And for wet basis, we have to substract the molar fraction of B <u>AND </u>the molar fraction of water vapor. So, logically, the mole fraction Xa will be less for wet basis.

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For the reaction of C 2H 4( g) with O 2( g), to form CO 2( g) and H 2O( g), what number of grams of CO 2 could be produced from
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4.58g of CO₂ could be produced

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Based on the reaction:

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<em>1 mole of C₂H₄ reacts with 3 moles of oxygen to produce 2 moles of CO₂</em>

<em />

To solve this question we must find the moles of each reactant in order to find limiting reactant. With limiting reactant we can find the moles -And the mass- of CO₂ produced:

<em>Moles C₂H₄ -Molar mass: 28.05g/mol-</em>

2.0g * (1mol / 28.05g) = 0.0713moles

<em>Moles O₂ -Molar mass: 32g/mol-</em>

5.0g * (1mol / 32g) = 0.156moles

For a complete reaction of 0.0713 moles of C2H4 are required:

0.0713 moles C₂H₄ * (3 moles O₂ / 1 mol C₂H₄) = 0.214 moles of O₂

As there are just 0.156 moles, O₂ is limiting reactant.

The moles of CO₂ produced are:

0.156 moles O₂ * (2mol CO₂ / 3mol O₂) = 0.104 moles CO₂

The mass is -Molar mass CO₂: 44.01g/mol-

0.104 moles CO₂ * (44.01g / mol) =

<h3>4.58g of CO₂ could be produced</h3>
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