# Answer to Question #21730 in Inorganic Chemistry for Kailey

Question #21730

In a particular reaction between copper metal and silver nitrate, 12.7 g Cu produced 38.1 g Ag. What is the percent yield of silver in this reaction?

Expert's answer

In this problem formula is:

actual/theoretical *100

so you have to find the theoretical :

moles of CU = grams/RMM = 12.7/63.55 = 0.200

moles of Ag = 0.200x2 (because the stiochemstry equations shows a 2:1ratio)= 0.4moles

grams = moles x Rmm = 0.4 x 107.9 = 43.16grams

so that's how we find the theoretical .. using the top equation

=(38.1/43.16)x100 = 88.39%

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