Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises - Page 116: 3.64c


$mass(Fe) = 244.62g$ $mass(CO_2) = 289.08g$

Work Step by Step

Balanced equation: (3.64a) $Fe_2O_3+3CO−−>2Fe+3CO_2$ 1. Convert the mass of (Fe2O3) to nº of moles: $mm(Fe_2O_3)=55.85∗2+16∗3=159.7g/mol$ $n(moles)=mass(g)*mm$ $n(moles)=350*159.7$ $n(moles)=2.19$ 2. Use the proportions of the balanced equation to find the number of moles of Fe and CO2 formed: Fe: $\frac{1}{2}=\frac{2.19}{x}$ $x=2∗2.19=4.38moles$ CO2: $\frac{1}{3}=\frac{2.19}{y}$ $y=3∗2.19=6.57moles$ 3. Convert these numbers to grams: Fe: $mm(Fe)=55.85g/mol$ $mass(g)=mm∗n(moles)$ $mass(g)=55.85∗4.38$ $mass(g)=244.62g$ CO2: $mm(CO2)=12∗1+16∗2=44g/mol$ $mass(g)=mm∗n(moles)$ $mass(g)=44∗6.57$ $mass(g)=289.08g$
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