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Last updated date: 04th Dec 2023
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# Solid calcium carbonate, $CaC{O_3}$, is able to remove sulphur dioxide from waste gases by the reaction (balanced as written): $CaC{O_3} + S{O_2} + {\text{Other reactants}} \to CaS{O_3} + {\text{Other products}}$ In a particular experiment, $255g$ of $CaC{O_3}$ was exposed to $135g$of $S{O_2}$ in the presence of an excess amount of the other chemicals required for the reaction. a) What is the theoretical yield of $CaS{O_3}$? b) If only $198g$ of $CaS{O_3}$ was isolated from the products. What was the percentage yield $CaS{O_3}$ in this experiment?

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Hint: We know that Synthetic responses in reality don't generally go precisely as moved toward paper. Over the span of an examination, numerous things will add to the development of fewer products than anticipated. Other than spills and other test mistakes, there are typically misfortunes because of an inadequate response, bothersome side responses, and so forth Physicists need an estimation that shows how fruitful a response has been. This estimation is known as the percent yield.

To process the percent yield, it is first important to decide the amount of the product ought to be shaped dependent on stoichiometry. This is known as the hypothetical yield, the most extreme measure of product that could be framed from the given measures of reactants. The genuine yield is the measure of a product that is really shaped when the response is done in the research center. The percent yield is the proportion of the genuine respect the hypothetical yield, communicated as a rate:
${\text{Percent yield}} = \dfrac{{{\text{Actual yield}}}}{{{\text{Theoretical yield}}}} \times 100\%$
Calculate the number of moles of calcium carbonate,
Moles of calcium carbonate$= \dfrac{{255g}}{{100g/mol}}$
The number of moles of calcium carbonate $= 2.55moles$.
Calculate the number of moles of sulfur dioxide,
${\text{Moles of }}S{O_2} = \dfrac{{133g}}{{64g/mol}} = 2.07moles$
The theoretical yield can be calculated as,
${\text{Theoretical yield}} = \left( {40 + 32(3 \times 16)} \right)2.11$
On simplification we get,
${\text{Theoretical yield}} = 253.2g$
The percentage yield is calculated as,
$\% yield = \dfrac{{198}}{{253.2}} \times 100$
On simplification we get,
$\% yield = 78.2\%$

Note:
We have to know that much time and cash is spent improving the percent yield for substance creation. At the point when complex synthetic compounds are blended by various responses, one stage with a low percent yield can rapidly cause an enormous misuse of reactants and superfluous cost.
Regularly, percent yields are naturally under $100\%$ as a result of the reasons demonstrated before. In any case, percent yields more noteworthy than $100\%$ are conceivable if the deliberate result of the response contains pollutants that cause its mass to be more prominent than it really would be if the item was unadulterated.