Question

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?

Answer 1

`66.67\text{ \%}`

Step-by-step explanation:

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

Mathematically:

`\text{ Percentage Yield = }\frac{Actual\text{ yield}}{Theoretical\text{ Yield}}\text{ }*\text{ 100 \%}`

From the question:

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

Substituting the values, we have it that:

`Percentage\text{ Yield = }(18)/(27)\text{ }*\text{ 100 \% =66.67 \%}`