Question

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A chemist reacted 11.50 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 85%, what is the actual yield? Show your work including, the use of stoichiometric calculations and conversion factors.

Answer 1

Answer:

24.86 g of sodium chloride

Explanation:

The balanced reaction equation is 2Na(s) + Cl2(g) --------> 2NaCl(s)

Since chlorine gas was said to be in excess in the question so sodium is the limiting reactant.

From the reaction equation:

46g of sodium metal yields 117g of sodium chloride

11.50g of sodium yields 11.50×117/46 = 29.25g of sodium chloride

Since % yield of sodium chloride= 85%

% yield= actual yield/ theoretical yield ×100

85= actual yield/29.25×100

Actual yield= 85×29.25/100

Actual yield= 24.86 g of sodium chloride

Answer 2

Since Chlorine is in excess, this is a limiting reagent problem.
1) convert 11.50 g Na to g of NaCl using the balanced equation.
2) percent yield = (actual yield)/(potential yield).

.85= (actual yield)/(g from step 1)

Solve for actual yield