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Calculating percent yield in a reaction

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A chemist ran the reaction CH3CH2Cl(l) + KOH(aq) ^ CH3CH2OH(l) +KCL(aq) by heating 47.2 grams of CH3CH2Cl with excess KOH and obtained a yield of 28.7 grams of CH3CH2OH. What was his % yield?

I do not know where to begin with this problem. It only gives grams of one reactant? What is the actual yield? Detailed step by step instructions please.

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Solution Preview

The mass of the given reagent indicates that it is the limiting reagent, and the one that you base the theoretical yield on.

Find moles of CH3CH2Cl m/MM and then since it ...

Solution Summary

This solution is provided in 141 words. It discusses how to find the percent yield using moles and the limiting reagent.

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