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27 mol Na reacts with 27 mol Cl2 to produce NaCl. if a student carries out this procedure and obtains 18 mol of the theoretical 27 mols of NaCl, what is the percent yeild?

Respuesta :

Answer:

[tex]66.67\text{ \%}[/tex]

Explanation:

Here, we want to get the percent yield of the reaction

Mathematically:

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

From the question:

The actual yield is 18 mol while the theoretical yield is 27 mol

Substituting the values, we have it that:

[tex]Percentage\text{ Yield = }\frac{18}{27}\text{ }\times\text{ 100 \% =66.67 \%}[/tex]

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