Asked by Lacey
Consider the following reaction:
Zn (s) + 2HCl (aq) ---> ZnCl2 (g) + H2 (g)
If the reaction has a 68.4% yield, how many grams of zinc are needed to obtain an actual yield of 1.00 g of zinc chloride?
Stuck on how to figure this out!
Zn (s) + 2HCl (aq) ---> ZnCl2 (g) + H2 (g)
If the reaction has a 68.4% yield, how many grams of zinc are needed to obtain an actual yield of 1.00 g of zinc chloride?
Stuck on how to figure this out!
Answers
Answered by
Damon
Zn = 65.4 grams/mol
Cl = 35.5 grams/mol
H = 1 gram/mol
so
2HCl = 2(36.5g/mol) = 73 g
ZnCl2 = 136.4 g/mol
to get 1 gram ZnCl2 need
(1/.684)(65.4/136.4)= .701 grams Zn
Cl = 35.5 grams/mol
H = 1 gram/mol
so
2HCl = 2(36.5g/mol) = 73 g
ZnCl2 = 136.4 g/mol
to get 1 gram ZnCl2 need
(1/.684)(65.4/136.4)= .701 grams Zn
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