Question

What mass of propene is formed from 34.0 g of iodopropane on heating with ethanolic $KOH$, if the yield is 36 %?

• A. 17.2 g
• B. 8.4 9
• C. 3.024 g
• D. 1.72 g

Answer 1

The correct option is C 3.024 g

$C{H}_{3}C{H}_{2}C{H}_{2}I+KOH⟶C{H}_3-CH=C{H}_2+KI+H_{2}O$

Moles of Iodopropane $=\frac{34}{170}=0.2$ $moles$

$1$ $mole$ of iodopropane gives $1$ $mole$ propene but yield is $36\%$.

So, $1$ $mole$ of iodopropane will give $0.36$ $mole$ of propene.

$0.2$ $mole$ of iodopropane will give $0.2\times 0.36=0.072$ $mole$ of propene.

Mass of propene $=42\times 0.072=3.024$ $gram$