Question

A chemist reacted 57.50 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is shown: Na + Cl2 → NaCl. If the percentage yield of the reaction is 86%, what is the actual yield?

Answer 1

To calculate the actual yield of NaCl, we first need to determine the theoretical yield. From the balanced equation, we can see that 2 moles of sodium react with 1 mole of chlorine gas to produce 2 moles of sodium chloride. The molar mass of sodium is 22.99 g/mol, so 57.50 g of sodium is equal to 2.50 mol Na.

Step-by-step explanation:

To calculate the actual yield of NaCl, we first need to determine the theoretical yield. The balanced equation for the reaction is:

2 Na (s) + Cl2(g) → 2 NaCl (s)

From the equation, we can see that 2 moles of sodium react with 1 mole of chlorine gas to produce 2 moles of sodium chloride.

The molar mass of sodium is 22.99 g/mol, so 57.50 g of sodium is equal to:

(57.50 g Na) / (22.99 g/mol Na) = 2.50 mol Na

Since the reaction has a 1:1 stoichiometry between sodium and sodium chloride, we can say that 2.50 mol Na will produce 2.50 mol NaCl. Using the molar mass of NaCl (58.44 g/mol), we can calculate the theoretical yield:
(2.50 mol NaCl) * (58.44 g/mol NaCl) = 145.1 g NaCl

Finally, to calculate the actual yield, we multiply the theoretical yield by the percentage yield:
(145.1 g NaCl) * (0.86) = 124.86 g NaCl