2A1 + Fe,O, → 2Fe + Al,O, AH =-845kJ 3 a) 50.0g Al x 1 mol Al 26.98g Al -845 k.J X 1 mol Al = -1570kJ b) 50.0g Al x I mol Al -845 kJ 2 mol Al = -783kJ 26.98g Al -845 kJ 2 mol Al c) 50.0g Al x = -21,100kJ d) 50.0g Al x 1 mol Al 26.98g Al 2 mol Al X -845 kJ = -0.00439kJ

Chemistry & Chemical Reactivity
9th Edition
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter5: Principles Of Chemical Reactivity: Energy And Chemical Reactions
Section5.5: Enthalpy Changes For Chemical Reactions
Problem 1RC: 1. For the reaction 2 Hg(l) + O2(g) → 2 HgO(s), ∆rH° = 181.6 kJ/mol-rxn. What is the enthalpy change...
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 Which of these shows the correct work to calculate the kJ of heat released if 50.0g of Al is reacted using the reaction below?

A

B

D

2Al + Fe,O, –→ 2Fe + Al,O,
AH = -845kJ
-845 kJ
1 mol Al
-845 kJ
a) 50.0g Al x Imol Al
= -1570kJ
X
26.98g Al
1 mol Al
b) 50.0g Al x
-783kJ
X
26.98g Al
-845 kJ
2 тol Al
c) 50.0g Al x
d) 50.0g Al x 26.98g Al
-21,100kJ
%D
2 mol Al
1 mol Al
2 mol Al
-845 kJ
-0.00439kJ
Transcribed Image Text:2Al + Fe,O, –→ 2Fe + Al,O, AH = -845kJ -845 kJ 1 mol Al -845 kJ a) 50.0g Al x Imol Al = -1570kJ X 26.98g Al 1 mol Al b) 50.0g Al x -783kJ X 26.98g Al -845 kJ 2 тol Al c) 50.0g Al x d) 50.0g Al x 26.98g Al -21,100kJ %D 2 mol Al 1 mol Al 2 mol Al -845 kJ -0.00439kJ
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