# chemistry

Given the following reaction:
Pd(NO3)2+CaSO4-->PdSO4+Ca(NO3)2
If you started with 225.6ml of 6.32M CaSO4 and the percent yield of this reaction was ol 78.2%, what is the actual yield of grams mod of PdSO4?

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1. Your next to last line doesn't make sense but I assume you want to know how many gram PdSO4 you have at the end of the reaction.
moles CaSO4 = M x L
Using the coefficients in the balanced equation, convert moles CaSO4 to moles PdSO4.
Now convert moles PdSO4 to grams.
g = moles x molar mass. This is the theoretical yield.
Then theoretical yield x 0.782 = g actual yield.

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posted by DrBob222

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