Areaction of 58.3 g of na and 54.1 g of br 2 yields 59.6 g of nabr . what is the percent yield? 2 na ( s ) + br 2 ( g ) ⟶ 2 nabr ( s )

The answer to your question is percent yield = 85.6

Explanation:

Data

mass of Na = 58.3 g

mass of Br = 54.1 g

mass of NaBr = 59.6 g

Reaction

                   2Na(s)  +  Br₂   ⇒   2NaBr (s)

Process

1.- Calculate the atomic mass of the reactants and products

sodium = 2 x 23 = 46 g

bromine = 2 x 80 = 160 g

NaBr = 2(23 + 80) = 206 g

2.- Calculate the limiting reactant using proportions

Theoretical       Na / Br = 46 / 160 = 0.2875

Experimental    Na / Br = 58.3 / 54.1 = 1.077

From these data we conclude that the limiting reactant is Br, because the proportion was higher in the experiment than means that there was less Bromine or more Sodium.

3.- Calculate the theoretical production of NaBr

                        160 g of Br --------------- 206 g of NaBr

                        54.1 g of Br ---------------  x

                        x = (54.1 x 206) / 160

                        x = 11144.6 / 160

                        x = 69.65 g of NaBr

4.- Calculate the percent yield

        % yield =

       % yield = 85.6

the answer is 303.15 or 303

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