Answer
0.59 g NO.
Work Step by Step
The strategy to find the mass of NO that is formed when we know the amount of the other product, 5.58 g Cu$(NO_{3})_2$ :
g Cu$(NO_{3})_2$ ${\rightarrow}$ mol Cu$(NO_{3})_2$ ${\rightarrow}$ mol NO ${\rightarrow}$ g NO
The conversion factors are:
$\frac{1 mol (Cu(NO_{3})_2)}{Molar mass (Cu(NO_{3})_2)}$ to convert mass to moles Cu$(NO_{3})_2$, where Molar mass Cu$(NO_{3})_2$:
Molar mass Cu$(NO_{3})_2$ = 1 x 64.0 + 2 x 14.0 + 6 x 16.0 = 188.0 g
$\frac{2 mol (NO)}{3 mol (Cu(NO_{3})_2)}$, to convert the mol of Cu$(NO_{3})_2$ to mol of NO.
$\frac{Molar mass(NO)}{1 mol(NO)}$ to convert to g of NO, where the molar mass of NO is:
Molar mass NO = 1x 14 +1 x 16.0 = 30.0 g
Now we use the conversion factors to find the mas of NO that is obtained, when we know the mass of 5.58 g of Cu$(NO_{3})_2$:
5.58 g Cu$(NO_{3})_2$ x $\frac{1 mol (Cu(NO_{3})_2)}{Molar mass (Cu(NO_{3})_2)}$ x $\frac{2 mol (NO)}{3 mol (Cu(NO_{3})_2)}$ x $\frac{Molar mass(NO)}{1 mol(NO)}$ =
5.58 g Cu$(NO_{3})_2$ x $\frac{1 mol (Cu(NO_{3})_2)}{188.0g (Cu(NO_{3})_2)}$ x $\frac{2 mol (NO)}{3 mol (Cu(NO_{3})_2)}$ x $\frac{30.0g (NO)}{1 mol(NO)}$ = 0.59 g NO