Question

If $27g$ of water is formed during complete combustion of pure propene $\left(C_3{H}_6\right)$, the mass of propene burnt is:

• A. $42g$
• B. $21g$
• C. $14g$
• D. $56g$
• E. $40g$

Answer 1

The correct option is B $21g$

Required reaction:
$\begin{array}{c}{C}_3{H}_6\\ \left(Propene\right)(3\times 12+6=42)\end{array}\begin{array}{c}+\\ \end{array}\begin{array}{c}\frac{9}{2}{O}_2\\ \end{array}\begin{array}{c}⟶\\ \end{array}\begin{array}{c}3C{O}_2\\ \end{array}\begin{array}{c}+\\ \end{array}\begin{array}{c}3H_{2}O\\ (3\times 18=54)\end{array}$
Given, $H_{2}O$ (formed) $=27g$
To find the mass of propene,
The mass ratio between $C_3{H}_6$ and $H_{2}O=42:54$
$\because 54g$ of $H_{2}O$ require $=42g$ $C_3{H}_6$
$\therefore 27g$ of $H_{2}O$ require $=\frac{42\times 27}{54}=21g\cdot C_3{H}_6$