Question

Below, make sure to SHOW ALL WORK to receive credit. If only an answer is provided you will receive a 0/10You are in a lab and you use 11.50 g of Na(s) to react with Cl(g). Below is the reaction that happens from this.Na + Cl2 → NaClYour percent yield is 95%. From this info what is the actual yield?

Answer 1

The actual yield is also the amount of product you obtained for that % yield

for that we have to calcualte first the theoretical yeild or in other words how mane grams of NaCl would be obtained if all 11.50 grams of Na are conberted into NaCl For that we need the molar mass of Na and NaCl

Mm of Na= 23 g/mol ; Mm of NaCl = 58.5 g/mol.

Which tht information and according to the stechiometry of the reaction we know that 23g of Na produce 58.5 g of NaCl. If we use this information as conversion factor and multipli it by the initial amount of Na we can calculate the theoretical yield:

`\text{theoretical yeild = 11.5g of Na}\frac{58.5\text{ g of NaCl}}{23\text{ g og Na}}=29.25\text{ g of NaCl}`

tho obtain the actual yield we have to multiply the theoretical yield times the percentage yield: