Asked by Anonymous
Zinc reacts with 387 mL of 4.00 M cold, aque-
ous sulfuric acid through single replacement.
a. How much zinc sulfate is produced?
Answer in units of g
b. How many liters of hydrogen gas would be
released at STP?
Answer in units of L
-- I have solved part a, but I am confused on part B. I would think that since there is only 1 mol of hydrogen gas produced, that at STP, there would be 22.4 L of H2 produced, but I know that is the incorrect answer. I would know how to solve B if I knew that the original information was also at STP. Can you assume that it is?
ous sulfuric acid through single replacement.
a. How much zinc sulfate is produced?
Answer in units of g
b. How many liters of hydrogen gas would be
released at STP?
Answer in units of L
-- I have solved part a, but I am confused on part B. I would think that since there is only 1 mol of hydrogen gas produced, that at STP, there would be 22.4 L of H2 produced, but I know that is the incorrect answer. I would know how to solve B if I knew that the original information was also at STP. Can you assume that it is?
Answers
Answered by
DrBob222
I would assume that since an amount for Zn is not given that H2SO4 is the limiting reagent (LR).
Zn + H2SO4 ==> ZnSO4 + H2
You have 4.00 x 0.387 = about 1.55 mols H2SO4 which will produce 1.55 mols H2 at STP. (Note that you don't have 1 mol H2 which is the error in your thinking.)
Since 1 mol of a gas at STP occupies 22.4L, 1.55 mols will occupy 22.4 x 1.55 = ? Remember 1.55 is an estimate; you calculate should confirm that.
Zn + H2SO4 ==> ZnSO4 + H2
You have 4.00 x 0.387 = about 1.55 mols H2SO4 which will produce 1.55 mols H2 at STP. (Note that you don't have 1 mol H2 which is the error in your thinking.)
Since 1 mol of a gas at STP occupies 22.4L, 1.55 mols will occupy 22.4 x 1.55 = ? Remember 1.55 is an estimate; you calculate should confirm that.
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