Asked by Anonymous
a 5.95 g sample of AgNO3 is reacted with excess BaCl2 according to the equation 2AgNO3 + BaCl2 yields 2AgCl + Ba(NO3)2 to give 3.17 g of AgCl. What is the percent yield of AgCl?
Answers
Answered by
DrBob222
2AgNO3 + BaCl2 ==> 2AgCl + Ba(NO3)2
mols AgNO3 = grams/molar mass
Using the coefficients in the balanced equation, convert mols AgNO3 to mols
AgCl.
g AgCl = mols AgCl x molar mass
AgCl = ?. This is the theoretical yield; i.e., the yield if it were 100%.
Then % yield = (3.17/theoret yield)*100 = ?
mols AgNO3 = grams/molar mass
Using the coefficients in the balanced equation, convert mols AgNO3 to mols
AgCl.
g AgCl = mols AgCl x molar mass
AgCl = ?. This is the theoretical yield; i.e., the yield if it were 100%.
Then % yield = (3.17/theoret yield)*100 = ?
Answered by
Anonymous
56
Answered by
Anonymous
63.1506
Answered by
Val
in a particular experiment to prepare a compound, the theoritical yield is 34.7 g. if the actual yield is 31. 2 g, what is the percentage yield?
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