## Chemistry: The Central Science (13th Edition)

Published by Prentice Hall

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

#### Answer

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

#### Work Step by Step

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

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