Question

2.42 mol of sodium is exposed to excess chlorine gas. Calculate the percent yield if 151g of table salt is collected at the end of the experiment.

Answer 1

The percent yield for the reaction is 106.77%

Step-by-step explanation

Given:

Moles of Na = 2.42 mol

Actual yield of table salt (NaCl) produced = 151 g

What to find:

The percent yield for the reaction.

Step-by-step solution:

The first step is to write a balanced equation for the reaction.

2Na + Cl₂ → 2NaCl

From the equation,

2 moles of Na produced 2 moles of NaCl

So, 2.42 moles of Na will produce x moles of NaCl

The value of x is calculated below

`x=\frac{2.42\text{ }mol\text{ }Na}{2\text{ }mol\text{ }Na}*2\text{ }mol\text{ }NaCl=2.42\text{ }mol\text{ }NaCl`

The next step is to calculate the theoretical yield of NaCl.

1 mole NaCl = 58.44 grams

2.42 moles NaCl = x grams

`x=(2.42mol)/(1mol)*58.44\text{ }grams=141.4248\text{ }g\text{ }NaCl`

The final step is to calculate the percent yield using the formula below:

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

Actual yield = 151 g

Theoretical yield = 141.4248 g

Thus,

`Percent\text{ }yield=(151g)/(141.4248g)*100\%=106.77\%`

The percent yield for the reaction is 106.77%