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

1. Convert that number to moles: $mm(O_2) = 16 * 2 = 32$ $n(moles) = \frac{mass(g)}{mm}$ $n(moles) = \frac{7.5}{32}$ $n(moles) \approx 0.234$ 2. Using the proportion given in the problem, find how much moles of $KO_2$ will react with 0.234 moles of $O_2$ $\frac{4}{3} = \frac{x}{0.234}$ $3x = 0.936$ $x = 0.312 moles$ 3. Convert that number to grams: $mm(KO_2) = 39.1*1 + 16*2 = 71.1 g/mol$ $mass(g) = n(moles) * mm$ $mass(g) = 0.312 * 71.1$ $mass(g) \approx 22.18g$