Answer
7.94 g $NO_{2}$
Work Step by Step
The strategy to find the mass of $NO_{2}$ that isformed when we know the amount of the other product 1.381 g $O{2}$ :
g $O_{2}$ ${\rightarrow}$ mol $O_{2}$ ${\rightarrow}$ mol $NO_{2}$${\rightarrow}$ g $NO_{2}$
The conversion factors are:
$\frac{1 mol (O_{2})}{Molar mass (O_{2})}$ to convert to moles $O_{2}$, where Molar mass $O_{2}$:
Molar mass $O{2}$ = 2 x 16.0 = 32.0 g
$\frac{4 mol (NO_{2})}{1 mol (O_{2})}$, to convert the moles of $O_{2}$ to mol of $NO_{2}$.
$\frac{Molar mass(NO_{2})}{1 mol(NO_{2})}$ to convert to g of $NO_{2}$, where the molar mass of $NO_{2}$ is:
Molar mass $NO_{2}$ = 1x 14 + 2 x 16.0 = 46.0 g
Now we use the conversion factors to find the mas of $NO_{2}$:
1.381 g $O_{2}$ x $\frac{1 mol (O_{2})}{Molar mass (O_{2})}$ x $\frac{4 mol (NO_{2})}{1 mol (O_{2})}$ x $\frac{Molar mass(NO_{2})}{1 mol(NO_{2})}$ =
1.381 g $O_{2}$ x $\frac{1 mol (O_{2})}{32 g (O_{2})}$ x $\frac{4 mol (NO_{2})}{1 mol (O_{2})}$ x $\frac{46 g(NO_{2})}{1 mol(NO_{2})}$ = 7.94 g $NO_{2}$
