Theoretical Yield Ni + 2HCl yields NiCl2 + H2

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Theoretical Yield Ni + 2HCl yields NiCl2 + H2

Post by George » Thu Mar 09, 2017 9:58 am

Hey everybody,

I have the balanced equation of Ni (s) + 2 HCl (aq) -> NiCl2 (aq) + H2 (g)

A.) If 12.0 moles of HCl were reacted, how many moles of Ni would be needed such that there was no excess?

B.) What is the theoretical yield (in grams) of NiCl2 from reacting 12.0 mol HCl?

C.) If 12.0 mol of HCl was reacted and the percent yield of the reaction was 78.5%, what was the actual yield (grams) of NiCl2?

A) Just use the coefficients of the balanced equation to relate moles of one thing to moles of another:
12 mol HCl X (1 mol Ni / 2 mol HCl) = 6 mol Ni

B) Again, use the coefficients from the equation to calculate moles NiCl2 formed. Then use the molar mass of NiCl2 to convert moles to grams:
12 mol HCl X (1 mol NiCl2 / 2 mol HCl) = 6 mol NiCl2
6 mol NiCl2 X (129.6 g/mol) = 778 grams NiCl2 (THIS is the theoretical yield for this reaction)

C) The percent yield is calculated as:
% yield = (actual yield / theoretical yield) X 100.

So, substituting what you know into this gives:

78.5% = (actual yield / 778 g) X 100
0.785 = actual yield / 778
actual yield = 611 grams NiCl2

Do these answers make sense?

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