Asked by Tsunayoshi
A student used 3.27g of Zn to react with excess HCl and obtains 6.47g ZnCl2. What is the percent yield?
Answers
Answered by
DrBob222
First, calculate theoretical yield (TY).
Zn + 2HCl ==> ZnCl2 + H2
mols Zn = grams/atomic mass = about 0.05 but that's an estimate.
Using the coefficients in the balanced equation, convert mols Zn to mols ZnCl2.
That's about 0.05 x (1 mol ZnCl2/1 mol Zn) = estd 0.05 mols ZnCl2.
g ZnCl2 = mols ZnCl2 x molar mass ZnCl2. This is the TY.
Actual yield (AY) is 6.47g.
%yield = (AY/TY)*100 = ?
Zn + 2HCl ==> ZnCl2 + H2
mols Zn = grams/atomic mass = about 0.05 but that's an estimate.
Using the coefficients in the balanced equation, convert mols Zn to mols ZnCl2.
That's about 0.05 x (1 mol ZnCl2/1 mol Zn) = estd 0.05 mols ZnCl2.
g ZnCl2 = mols ZnCl2 x molar mass ZnCl2. This is the TY.
Actual yield (AY) is 6.47g.
%yield = (AY/TY)*100 = ?
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