Question

What mass of propene is formed from 34.0 g of 1-iodopropane on treating with ethanolic $KOH$, which yields 36%?

• A. $6.048g$
• B. $4.12g$
• C. $3.024g$
• D. $5.98g$

Answer 1

The correct option is C $3.024g$

$C{H}_3-C{H}_2-C{H}_2-I\underset{-KI}{\overset{\text{ethanolic KOH}}{\to }}C{H}_3-CH=C{H}_2$

The molecular weights of 1-iodopropane and propene are 170 g/mol and 42 g/mol respectively.

Number of moles of 1 - iodopropane $=\frac{34.0g}{170g/mol}=0.2mol$

Number of moles of propene $=$ number of moles of 1-iodopropane $=0.2mol$

Mass of propane $=0.2mol\times 42g/mol=8.4g$

This is assuming 100% yield. But actual yield is only 36 %

Mass of propane $=8.4g\times \frac{36}{100}=3.024g$