Question

$Ca{C}_2+2D_{2}O\to C_2{D}_2+Ca(OD{)}_2$ ...(i)
$S{O}_3+D_{2}O\to D_{2}S{O}_4$ ...(ii)
$A{l}_4{C}_3+12D_{2}O\to 3C{D}_4+4Al(OD{)}_3$ ...(iii)
$C_2{D}_2+O_2\to C{O}_2+D_{2}O$ ...(iv)
$C{D}_4+O_2\to C{O}_2+D_{2}O$ ...(v)
If $Ca{C}_2,S{O}_3\&A{l}_4{C}_3$ are taken in $3:2:1$ molar ratio and hydrocarbon obtained is further treated with $O_2$. Then the ratio of mass of $D_{2}O$ consumed in the first three steps to the mass of $D_{2}O$ formed in last two steps are

• A. $\frac{6}{19}$
• B. $\frac{20}{9}$
• C. $3$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{20}{9}$

Answer (b)
$\underset{3n}{Ca{C}_2}+\underset{6n}{2D_{2}O}\to \underset{3n}{C_2{D}_2}+Ca{\left(OD\right)}_2$
$\underset{2n}{S{O}_3}+\underset{2n}{D_{2}O}\to D_{2}S{O}_4$
$\underset{n}{A{l}_4{C}_3}+\underset{12n}{12D_{2}O}\to \underset{3n}{3C{D}_4}+4Al{\left(OD\right)}_3$
$D_{2}O$ consumed $=6n+2n+12n=20n\text{mole}$
$\underset{3n}{C_2{D}_2}+\frac{5}{2}{O}_2\to 2C{O}_2+\underset{3n}{D_{2}O}$
$\underset{3n}{C{D}_4}+2O_2\to C{O}_2+2\underset{6n}{D_{2}O}$
$D_{2}O$ formed $=3n+6n=9n\text{mol}$
$\text{Ratio of weight}=\frac{20n\left(20\right)}{9n\left(20\right)}=\frac{20}{9}$