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The reaction of ethane gas (C2H6) with chlorine gas produces C2H5Cl as its main product (along with HCl). In addition, the reaction invariably produces a variety of other minor products, including C2H4Cl2, C2H3Cl3, and others. Naturally, the production of these minor products reduces the yield of the main product. Calculate the percent yield of C2H5Cl if the reaction of 300. g of ethane with 650. g of chlorine produced 490. g of C2H5Cl.

Respuesta :

Answer:

The percent yield of  chloro-ethane in the reaction is 82.98%.

Explanation:

[tex]C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl[/tex]

Moles of ethane = [tex]\frac{300.0 g}{30 g/mol}=10 mol[/tex]

Moles of chlorine gases =[tex]\frac{650.0 g}{71 .0 g/mol}=9.1549 mol[/tex]

As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.

This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.

According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.

Then 9.1549 moles of chlorien gas will give:

[tex]\frac{1}{1}\times 9.1549 mol=9.1549 mol[/tex] of chloro-ethane

Mass of 9.1549 moles of chloro-ethane:

9.1549 mol × 64.5 g/mol = 590.4910 g

Theoretical yield of  chloro-ethane: 590.4910 g

Given experimental yield of chloro-ethane: 490.0 g

[tex]\% Yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

[tex]\%Yield (C_2H_5Cl)=\frac{490.0 g}{590.4910 g}\times 100=82.98\%[/tex]

The percent yield of  chloro-ethane in the reaction is 82.98%.

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