Answer
Approximately 22.18 grams.
Work Step by Step
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:
$mm(KO_2) = 39.1*1 + 16*2 = 71.1 g/mol$
$mass(g) = n(moles) * mm$
$mass(g) = 0.312 * 71.1$
$mass(g) \approx 22.18g$
