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

Published by Prentice Hall

# Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises - Page 116: 3.62c

#### Answer

Approximately 6.86g.

#### Work Step by Step

1. Convert the mass of $O_2$ to nº of moles: $mm(O_2) = 2*16 = 32g/mol$ $n(moles) = \frac{mass(g)}{mm}$ $n(moles) = \frac{7.5}{32}$ $n(moles) = 0.234$ 2. Using the proportion gave in the balanced equation: $\frac{2}{3} = \frac{x}{0.234}$ $3x = 0.234 * 2$ $x = \frac{0.468}{3}$ $x = 0.156 moles (CO_2)$ 3. Convert that number to grams: $mm(CO_2) = 12*1 + 16*2 = 44g/mol$ $mass(g) = n(moles) * mm$ $mass(g) = 0.156 * 44$ $mass (g) = 6.864$

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