PLEASE SHOW ALL WORK! 1. Jesse has a 30.6 g sample of zinc that he wants to reac

Question

PLEASE SHOW ALL WORK! 1. Jesse has a 30.6 g sample of zinc that he wants to react with sulfuric acid, according to the following equation. H2SO4(aq) + Zn(s)- ZnSO4(aq)+ H2(g) a. Write the net ionic equation for this reaction. b. In this reaction, what is reduced? What is oxidized? (Note: You are given this much space for a reason. You must show work to justify your answer) c. If Jesse uses a 0.36 M H2SO4 solution, what volume of solution will he need to completely react with the zinc sample? d. If Jesse performs this reaction under standard conditions (25 C and 1.00 atm), what volume of hydrogen gas will be produced by the complete reaction of this zinc sample?

Explanation / Answer

A)H2SO4 (aq)+  Zn (s) ----------> ZnSO4 (aq)+ H2 (g)

Ionic Equation: H+(aq)+ SO42- (aq)+Zn (s)   ----------> Zn2+ aq)+ SO42-(aq)+ H2 (g)

Cancelling the common terms we get the net ionic equation:

2H+(aq)+Zn (s)   ----------> Zn2+ aq)+ H2 (g)

B) 2H+1(aq)+Zn0 (s)   ----------> Zn2+ aq)+ H20(g)

H is reduced as its charge went from +1 to 0

Zn is oxidized as its Oxidation state changed from 0 to +2.

C) Mass of Zn sample = 30.6 gms

Molar mass of Zn = 65.38 g/mol

moles of Zn = mass/ molar mass = 30.6/65.38 = 0.468 moles of Zn

For 0.468 moles of Zn, 0.468 moles of H2SO4 is required.

Volume of H2SO4 required = moles/molarity = 0.468 moles /0.36 moles/L = 1.30 L

D) PH2VH2 = nH2 R T

VH2 = (nH2 R T)/PH2

T = 25 deg c = 25 + 273 K = 298 K.

PH2= total pressure - pressure of water vapor at 25 deg = 1 atm - 23.8 torr.

= 1 atm -23.8/760 atm = 1atm - 0.0313 atm = 0.969 atm

R = 8.314 J/mol.K

nH2 = moles of Zn or moles of H2SO4 = 0.468 moles

VH2 = (0.468 moles * 0.08205 L.atm/mol.K * 298.15 K)/0.687 = 16.66 L

The total volume of H2 gas produced = 16.66 L

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