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1 year ago

In an experiment, 170.9 g of C2H4 was reacted with an excess of O2, 164.1 g of CO2 is produced.

Chemistry

1 answer:

jonny [76]1 year ago

7 0

Answer:

$Y=30.6\%$

Explanation:

Hello,

In this case, given the reaction, the molar mass of ethene is 28 g/mol and the molar mass of carbon dioxide is 44 g/mol. With that information we compute the theoretical yield considering a 1:2 molar ratio respectively between them:

$m_{CO_20}^{theoretical}=170.9gC_2H_4*\frac{1molC_2H_4}{28gC_2H_4}*\frac{2molCO_2}{1molC_2H_4} *\frac{44gCO_2}{1molCO_2} =537.1gCO_2$

Thus, we compute the percent yield with the given grams of carbon dioxide:

$Y=\frac{m_{CO_2}^{real}}{m_{CO_2}^{theoretical}}*100 \% =\frac{164.1g}{537.1gCO_2} *100 \%\\\\Y=30.6\%$

Regards.

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For a single component system, why do the allotropes stable at high temperatures have higher enthalpies than allotropes stable a

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2 years ago

15. An apparatus consists of a 4.0 dm3

Solnce55 [7]

Answer:

i· Partial pressure of nitrogen gas is 219.429kPa and partial pressure of argon gas is 33.714kPa ·

ii· Total pressure of the gas mixture is 253.143kPa·

Explanation:

<h3>solution for i :</h3>

Assuming that the given gases to be ideal,

so by ideal gas equation

PV=nRT

where,

P is the pressure of the gas

V is the volume occupied by the gas

n is the number of moles of the gas

R is the ideal gas constant

T is the temperature of the gas

Actually partial pressure of each gas is the pressure of the gas exerted when it occupies complete volume

As the temperature of both gases are same, the mixing process is an isothermal process

PV=constant

Initially for nitrogen gas PV=803×4=3212

let P $x_{1}$ be the partial pressure of the nitrogen gas

(P $x_{1}$ )×14=3212

∴P $x_{1}$ =219.429kPa

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Initially for argon gas PV=47.2×10=472

let P $x_{2}$ be the partial pressure of the argon gas

(P $x_{2}$ )×14=472

∴P $x_{2}$ =33.714kPa

∴Partial pressure of argon gas is 33.714kPa

solution for ii :

Total pressure of the gas mixture will be the sum of the partial pressures of each gas as

Partial pressure of the gas=(total pressure of the mixture)×(mole fraction of the gas)

∴Total pressure=P $x_{1}$ +P $x_{2}$

=219.429+33.719

Total pressure =253.143kPa

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