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riverside07 riverside07
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11 years ago
5.00 g of HCl are allowed to react with 25.0 g of zinc metal in the presence of water. During the reaction, zinc(II) chloride and hydrogen gas result. After the reaction is finished, [Zn2+(aq)] = 0.221 M, and the volume of the solution is 56.0 mL. What is the apparent percent yield of the reaction?
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wrote...
11 years ago
moles Zn = 25.0 g/ 65.39 g/mol=0.382
Zn + 2 HCl = ZnCl2 + H2
moles HCl required = 0.382 x 2 =0.764
moles HCl = 5.00 g /36.461 g/mol=0.137
HCl is the limiting reactant
moles Zn2+ formed = 0.137/2=0.0686
[Zn2+]= 0.0686/ 0.0560 L=1.22 M ( theoretical concentration)
% = 0.221 x 100/ 1.22 = 18.1
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