   Chapter 13.10, Problem 13.11SC ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
3 views

# trong>Exercise 13.11 Calculate the volume of hydrogen produced at 1 . 5 0  atm and 19 ° C by the reaction of 26 . 5 g of zinc with excess hydrochloric acid according to the balanced equationmsp;  Zn ( s )   +  2HCl ( a q )   → ZnCl 2 ( a q )   +  H 2 ( g )

Interpretation Introduction

Interpretation:

The volume of hydrogen gas produced at 1.5 atm and 19° C by the reaction of 26.5 g of zinc with excess hydrochloric acid should be determined according to the following balanced equation

Zn(s) + 2HCl(aq) ? ZnCl2 (aq) + H2 (g)

Concept Introduction:

Amount of hydrogen gas that produced from the reaction of 26.5 g of zinc with excess of HCl can be determined using the stoichiometry of Zn and H2 in the balanced equation. Once the amount of hydrogen gas is determined, the volume of hydrogen gas can be found using the ideal gas law. Ideal gas law is expressed as

PV = nRT

Where

P is the pressure of the gas

V is the volume of the gas (need to be determined)

T is the absolute temperature of the gas

n is the number of moles of the hydrogen gas molecule and

R is the gas constant.

Explanation

Given:

P = 1.5 atm

T = 19° C = (273+19) K = 292 K

From the balanced equation between the reaction of Zn and the excess of HCl, it is found that one mole of hydrogen gas is produced from 1 mole of Zn

Now, amount of Zn(s) used in the reaction is 26.5 g

Therefore, moles of Zn used = (26.5 g/65.38 gmol-1) = 0.405 mol

So, 0.405 moles of Zn(s) is used in the reaction, therefore amount of hydrogen gas produced is also 0...

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