Question

For the following reaction, if 32 g of Na, is reacted with excess Cl2 in the laboratory, and 56.9 g of NaCl is produced, what is the percentage yield of NaCl? Round your answer to the nearest whole percent.2 Na + Cl2 → 2 NaClSelect one:a.24 %b.99 %c.70 %d.86 %

Answer 1

The first step to answer this question is to convert the given mass of Na to moles:

$32gNa\cdot\frac{1molNa}{22.98gNa}=1.40molNa$

Using the stoichiometric ratio, we know that 2 moles of Na produce 2 moles of NaCl:

$1.40molNa\cdot\frac{2molNaCl}{2molNa}=1.40molNaCl$

Use the molecular weight of NaCl to convert the number of moles to grams:

$1.40molNaCl\cdot\frac{58.44gNaCl}{1molNaCl}=81.82gNaCl$

Now, divide the actual yield of NaCl by the theoretical yield of NaCl and multiply it by 100:

$yield=\frac{56.9gNaCl}{81.82gNaCl}\cdot100=70\%$

It means that the correct answer is c. 70%.