Question

A reaction of 55.1 g of Na and 31.1 g of Br2 yields 35.4 g of NaBr. What is the percent yield?

Answer 1

So,

% yield equals actual mass of product divided by theoretical mass of product times 100%.

1. First, write the balanced reaction.

$Na(s)+\frac{1}{2}Br_2(l) -->NaBr(s)$

2. Since we are given the masses of both reactants, we need to do a Limiting Reactant calculation.  Calculate the theoretical masses of product produced by each reactant, assuming excess of the other reactant.

a. $\frac{55.1g\ Na}{22.99g/mol}\ *\frac{1mol\ NaBr}{1mol\ Na}\ =2.40mol\ NaBr$

b. $\frac{31.1g\ Br_2}{159.8g/mol}\ *\frac{1mol\ NaBr}{0.5mol\ Br_2}\ =0.389mol\ NaBr$

c. Bromine is the limiting reactant, and the theoretical mass of NaBr is:

$0.389mol\ NaBr\ *102.89g/mol\ =40.0 g\ NaBr$

3. Now, we can calculate % yield.

$percent\ yield=\frac{mass_{actual}}{mass_{theoretical}}*100=\frac{35.4g}{40.0g}*100=88.5\ percent$

So, the percent yield is 88.5%.

Hope this helps!