Answer
12.9 g $O_{2}$ or 13 g $O_{2}$
Work Step by Step
The strategy to find the mass of $O_{2}$ that react with 5.5 g $NH_{3}$ is:
g $NH_{3}$ ${\rightarrow}$ mol $NH_{3}$ ${\rightarrow}$ mol $O_{2}$${\rightarrow}$ g $O_{2}$
The conversion factors are:
$\frac{1 mol (NH_{3})}{Molar mass (NH_{3})}$ to convert to moles $NH_{3}$, where Molar mass $NH_{3}$:
Molar mass $NH_{3}$ = 1 x 14.0 + 3 x 1.0 = 17.0 g
$\frac{5 mol (O_{2})}{4 mol (NH_{3})}$, to convert the moles of $NH_{3}$ to mol of $O_{2}$.
$\frac{Molar mass(O_{2})}{1 mol(O_{2})}$ to convert to g of $O_{2}$, where the molar mass of $O_{2}$ is:
Molar mass $O_{2}$ = 2 x 16.0 = 32.0 g
Now we use the conversion factors to find the mas of oxygen:
5.5 g $NH_{3}$ x $\frac{1 mol (NH_{3})}{Molar mass (NH_{3})}$ x $\frac{5 mol (O_{2})}{4 mol (NH_{3})}$ x $\frac{Molar mass(O_{2})}{1 mol(O_{2})}$ =
5.5 g $NH_{3}$ x $\frac{1 mol (NH_{3})}{17 g (NH_{3})}$ x $\frac{5 mol (O_{2})}{4 mol (NH_{3})}$ x $\frac{ 32 g(O_{2})}{1 mol(O_{2})}$ = 12.9 g $O_{2}$ or 13 g $O_{2}$
