I DONT WANT THE ANSWER. just confused on the steps and process. A chemist reacted 17.25 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is shown. Na + Cl2 → NaCl

If the percentage yield of the reaction is 88%, what is the actual yield? Show your work, including the use of stoichiometric calculations and conversion factors.

Na +Cl₂ → NaCl

First balance the given equation. The equation needs to be balanced because the number of chlorine atoms on the left side of the arrow is not equivalent to the number of Cl atoms of the right side of the arrow. After balancing the Cl atoms, ensure that the number of Na atoms are balanced on both sides of the arrow too.

The balanced equation would be:

2Na +Cl₂ → 2NaCl

Given the mass of Na metal, find the number of moles of Na metal using moles= mass ÷mr, where mr is the relative molecular mass of Na. This value can be obtained from the periodic table. Here, we are reacting Na metal with excess chlorine gas. This means that Na metal is the limiting reagent, thus the amount of NaCl obtained is dependent on the amount of Na metal present.

From the balanced equation, the mole ratio of Na to NaCl is 1: 1. Thus, if we have x moles of Na, we would obtain x moles of NaCl too. This amount of NaCl produced by the reaction is the theoretical yield.

However, in reality, the percentage yield is never 100% since there might be a loss of product during the reaction. Once you have found the mass of NaCl produced (let this value be y grams), find the actual yield.

☆ Mass= number of moles ×mr

100% yield y grams

1% yield (y/100) grams

88% yield 0.88y grams

Actual yield= 0.88y grams