Question

Match the reactions given in column I with the volumes given in column II

Answer 1

(A) $M=\frac{\text{percentage by weight}\times 10\times d}{M{w}_2}=\frac{90\times 10\times 1.8}{98}=18M$

$18M\times V_L=3.0M\times 3L\Rightarrow V_L=0.5L=500mL$
Thus, $500ml$ of $0.5M$ $HBr$ is required to meutralise $250mL0.2MBa(OH{)}_2$ solution.

$2HBr+Ba(OH{)}_2\to BaB{r}_2+2H_{2}O$

(B) Moles of $NaHC{O}_3=\frac{21.0}{84}=0.25mol$

Moles of $HCl=$ Moles of $NaHC{O}_3$
$3M\times V_L=0.25mol;V_L=0.083L=83.3mL$
$83.3mL$ of $2M{H}_{2}S{O}_4$ is required to produce $3.4g$ of $H_{2}S$ by the reaction. $8KI+5H_{2}S{O}_4\to K_{2}S{O}_4+4I_2+H_{2}S+4H_{2}O$

(C) Moles of $Zn=\frac{6.54}{65.4}=0.1mol$
Moles of $H_{2}S{O}_4=$ moles of $Zn$
$3M\times V_L=0.1,V_L=0.033L=33.3mL$
$33.3mL$ of $8.0M{H}_{2}S{O}_4$ is needed to react completely with 108 g of $Al$ (Atomic weight $Al=27.0g\left)\right[2Al+3H_{2}S{O}_4\to A{l}_2(S{O}_4{)}_3+H_2]$

(D) Moles of $K_2{C}_2{O}_4\cdot H_{2}O=\frac{92}{184}=0.5mol$
$5mol$ of $C_2{O}_4^{2-}\equiv 2$ mol of $Mn{O}_4^{\ominus }$
$0.5mol$ of $C_2{O}_4^{2-}\equiv 0.2$ mol of $Mn{O}_4^{\ominus }$
$M\times V_L$ of $Mn{O}_4^{\ominus }\equiv 0.2$ mol of $Mn{O}_4^{\ominus }$
$0.5\times V_L=0.2\Rightarrow V_L=0.4L=400mL$
$400mL$ of $1MBaC{l}_2$ is required to precipitate all the $S{O}_4^{2-}$ ions from $100mL$ of a solution containing $142gN{a}_{2}S{O}_4(Mw=142g)$ $BaC{l}_2+N{a}_{2}S{O}_4\to BaS{O}_4+3NaCl$