34 grams Iodo propane is treated with alcoholic KOH. What mass of propene is obtained if yield of the product is 50%(AW of Iodine=127)

$$\begin{gathered} {\mathrm{C}}_3{\mathrm{H}}_7\mathrm{I} + \mathrm{KOH}\left(\mathrm{alc}\right)\to {\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2+\mathrm{KI}+{\mathrm{H}}_2\mathrm{O}; \\ 170 \mathrm{g} 42 \mathrm{g} \\ 34 \mathrm{g} ? \\ \mathrm{Propene} \mathrm{formed} \mathrm{when} \mathrm{the} \mathrm{yield} \mathrm{is} 100\%= \frac{42\times 34}{170}=8.4\mathrm{gm}; \\ \mathrm{Since} \mathrm{the} \mathrm{yield} \mathrm{is} 50\% , \mathrm{propene} \mathrm{formed} =8.4\times 0.5=4.2 \mathrm{gm} \end{gathered}$$