LIST- I contains compounds and LIST- II contains reactions LIST- I LIST- II I H 2 O 2 P Mg HCO 3 2 + Ca OH 2 → II Mg OH 2 Q BaO 2 + H 2 SO 4 → III BaCl 2 R C a ( O H ) 2 + M g C l 2 → IV CaCO 3 S

Question

LIST- I contains compounds and LIST- II contains reactions LIST- I LIST- II I H 2 O 2 P Mg HCO 3 2 + Ca OH 2 → II Mg OH 2 Q BaO 2 + H 2 SO 4 → III BaCl 2 R C a ( O H ) 2 ​ + M g C l 2 ​ → IV CaCO 3 S BaO 2 + HCl → T Ca HCO 3 2 + Ca OH 2 → Match each compound in LIST- I with its formation reaction(s) in LIST- II , and choose the correct option

TestHub · Accepted Answer

In this method calculated amount of lime is added to hard water. It precipitates out calcium carbonate and magnesium hydroxide which can be filtered off $(P) Mg {({HCO}_{3})}_{2} + 2 Ca {(OH)}_{2} ⟶ 2 {CaCO}_{3} ↓ + Mg {(OH)}_{2} ↓ + 2 H_{2} O$
$(T) Ca {({HCO}_{3})}_{2} + Ca {(OH)}_{2} ⟶ 2 {CaCO}_{3} ↓ + 2 H_{2} O$
$(R) Ca {(OH)}_{2} + {MgCl}_{2} ⟶ Mg {(OH)}_{2} + {CaCl}_{2}$
Acidifying barium peroxide and removing excess water by evaporation under reduced pressure gives hydrogen peroxide
$(Q) {BaO}_{2} + H_{2} {SO}_{4} ⟶ {BaSO}_{4} + H_{2} O_{2}$
$(S) {BaO}_{2} + 2 HCl ⟶ {BaCl}_{2} + H_{2} O_{2}$
$I \to Q$
$II \to R$
$III \to S$
$IV \to P$

	LIST- $I$		LIST- $II$
$(I)$	$H_{2} O_{2}$	$(P)$	$Mg {({HCO}_{3})}_{2} + Ca {(OH)}_{2} \to$
$(II)$	$Mg {(OH)}_{2}$	$(Q)$	${BaO}_{2} + H_{2} {SO}_{4} \to$
$(III)$	${BaCl}_{2}$	$(R)$	$C a (O H)_{2} + M g C l_{2} \to$
$(IV)$	${CaCO}_{3}$	$(S)$	${BaO}_{2} + HCl \to$
		$(T)$	$Ca {({HCO}_{3})}_{2} + Ca {(OH)}_{2} \to$

Chemistry - Complete s-block Question with Solution | TestHub

Options:

Doubts & Discussion