CaC2+2D2O→C2D2+Ca(OD)2 ...(i)SO3+D2O→D2SO4 ...(ii)Al4C3+12D2O→3CD4+4Al(OD)3 ...(iii)C2D2+O2→CO2+D2O ...(iv)CD4+O2→CO2+D2O ...(v)If CaC2,SO3&Al4C3 are taken in 3:2:1 molar ratio and hydrocarbon obtained is further treated with O2. Then the ratio of mass of D2O consumed in the first three steps to the mass of D2O formed in last two steps are

CaC2+2D2O→C2D2+Ca(OD)2 ...(i)SO3+D2O→D2SO4 ...(ii)Al4C3+12D2O�

1.	CaC2+2D2O→C2D2+Ca(OD)2 ...(i)SO3+D2O→D2SO4 ...(ii)Al4C3+12D2O→3CD4+4Al(OD)3 ...(iii)C2D2+O2→CO2+D2O ...(iv)CD4+O2→CO2+D2O ...(v)If CaC2,SO3&Al4C3 are taken in 3:2:1 molar ratio and hydrocarbon obtained is further treated with O2. Then the ratio of mass of D2O consumed in the first three steps to the mass of D2O formed in last two steps are
Answer» $C a C_{2} + 2 D_{2} O \to C_{2} D_{2} + C a (O D)_{2}$ ...(i) $S O_{3} + D_{2} O \to D_{2} S O_{4}$ ...(ii) $A l_{4} C_{3} + 12 D_{2} O \to 3 C D_{4} + 4 A l (O D)_{3}$ ...(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 $C a C_{2}, S O_{3} & A l_{4} C_{3}$ 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