- kyonium

# Percentage Yield

Updated: Dec 7, 2020

Percentage yield is a sub chapter of chemical calculations. This article is meant to clarify this concept so that students will have no problems answering questions from this sub-chapter.
In an experiment, 23g of Na reacted with **excess** hydrochloric acid producing** 20 dm3** of hydrogen gas at room temperature and pressure. The reaction is as below,
Na + 2HCl --> NaCl + H2
Molar mass of Na is 23g
Mole of sodium present = 23/23 = 1 mol
From the equation, 1 mol of sodium will produce 1 mol of hydrogen gas. At room temperature and pressure, 1 mol of hydrogen gas will occupy a volume of 24 dm3.
However, in the experiment only 20 dm3 of hydrogen gas was collected , this means that the experiment did not produce 100% of the amount of hydrogen that we are expecting to get, in fact we are only getting 83% (20/24 x 100) of the expected amount , so the percentage yield is 83%.
So percentage yield = Actual Yield/Theoretical Yield x 100
__Summary: __
To find out the percentage yield, we need to
(1): Use the **chemical equation** to determine the theoretical amount of product (theorectical yield) obtained for the amount of reactant.
(2): Then we can used the given yield provided by the question to determine the percentage yield.
A high percentage yield implies that we are getting the maximum amount of product we can get for the amount of reactant.

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