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Chapter 9: Problem 61

Consider the unbalanced chemical equation \(\mathrm{HCl}+\mathrm{Zn} \rightarrow \mathrm{H}_{2}+\mathrm{ZnCl}_{2}\) (a) Balance the equation. (b) Translate the equation into words using the word mole \((s)\) wherever you can. (c) To produce 1 mole of \(\mathrm{H}_{2}\) from the reaction you just wrote, how many grams of \(\mathrm{HCl}\) and Zn must you combine? (d) What is the theoretical yield in grams of \(\mathrm{H}_{2}\) ? (e) You recover \(2.00 \mathrm{~g}\) of \(\mathrm{H}_{2}\) after carrying out the reaction. What is the percent yield?

Short Answer

Expert verified
(a) The balanced equation is: \(2 \mathrm{HCl} + \mathrm{Zn} \rightarrow \mathrm{H}_{2} + \mathrm{ZnCl}_{2}\) (b) Two moles of hydrochloric acid (HCl) react with one mole of zinc (Zn) to produce one mole of hydrogen gas (H₂) and one mole of zinc chloride (ZnCl₂). (c) To produce 1 mole of H₂, combine 72.9 g of HCl and 65.38 g of Zn. (d) The theoretical yield of H₂ is 2.00 g. (e) The percent yield of this reaction is 100%.

Step by step solution

01

Balance the given chemical equation

To balance the equation, we must equalize the number of each element's atoms on both sides of the equation. The balanced equation is: \[2 \mathrm{HCl} + \mathrm{Zn} \rightarrow \mathrm{H}_{2} + \mathrm{ZnCl}_{2}\]
02

Translate the balanced equation into words using "mole"

Two moles of hydrochloric acid (HCl) react with one mole of zinc (Zn) to produce one mole of hydrogen gas (H₂) and one mole of zinc chloride (ZnCl₂).
03

Calculate the required masses of HCl and Zn

To produce 1 mole of H₂, we need 2 moles of HCl and 1 mole of Zn. Mass of HCl = (2 moles) × (molecular weight of HCl) Molecular weight of HCl = 1 (H) + 35.45 (Cl) = 36.45 g/mol Mass of HCl = (2 moles) × (36.45 g/mol) = 72.9 g Mass of Zn = (1 mole) × (molecular weight of Zn) Molecular weight of Zn = 65.38 g/mol Mass of Zn = (1 mole) × (65.38 g/mol) = 65.38 g We need to combine 72.9 g of HCl and 65.38 g of Zn to produce 1 mole of H₂.
04

Calculate the theoretical yield of H₂ in grams

Theoretical yield of H₂ = (1 mole of H₂) × (molecular weight of H₂) Molecular weight of H₂ = 2(H) = 2 g/mol Theoretical yield of H₂ = (1 mole) × (2 g/mol) = 2.00 g
05

Calculate the percent yield

Percent yield = (Actual yield / Theoretical yield) x 100 Actual yield = 2.00 g (given in the problem) Percent yield = (2.00 g / 2.00 g) x 100 = 100% The percent yield of this reaction is 100%.

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