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(1molNa)/(22.98gNa)=1.40molNa`

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

`1.40molNa\cdot(2molNaCl)/(2molNa)=1.40molNaCl`

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

`1.40molNaCl\cdot(58.44gNaCl)/(1molNaCl)=81.82gNaCl`

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

`yield=(56.9gNaCl)/(81.82gNaCl)\cdot100=70\%`

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