Question

In a particular reaction between copper metal and silver nitrate, 12.7 g Cu produced 40.1 g of Ag. What is the percent yield of silver in this reaction? Cu(s) + 2AgNO3(aq) --> Cu(NO3)2(aq) + 2 Ag(s)

Answer 1

Answer:

b

Explanation:

b

Answer 2

Answer:

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Explanation:

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&039;s how we find the theoretical .. using the top equation

=(40.1/43.16)x100 = 92.91%