How many grams of CH_4 are present in 5.14*10^23 molecules of CH_4?

Well, $6.022 \times {10}^{23}$ methane molecules have a mass of $16.01 \cdot g$.
$\left(5.14 \times {10}^{23} \cdot {\text{methane molecules")/(6.022xx10^23*"methane molecules mol}}^{-} 1\right) \times 16.01 \cdot g \cdot m o {l}^{-} 1$
$\cong$ $13 \cdot g$