$(a)\;S\qquad(b)\;Ca\qquad(c)\;Fe\qquad(d)\;Na$

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$BaO \quad\underrightarrow {\Delta}\quad$ No effect

$MCO_3 \quad\underrightarrow {\Delta}\quad MO+CO_2$

$a+b=4.08$------(1)

Loss in weight=4.08-3.64=0.44g

Weight of $CO_2=0.44g$

Also weight of $CO_2$ given by 'b' g $MCO_3$

$\Rightarrow \large\frac{44b}{m+60}$

$\large\frac{44b}{m+60}$$=0.44$------(2)

Meq of $BaO$+Meq of $MCO_3$=Meq of BaO+Meq of Mo

$\Rightarrow $Meq of HCl used for those oxides

$\Rightarrow 100\times 1-(16\times 2.5)=60$

$\large\frac{a}{\Large\frac{153}{2}}$$\times 1000+\frac{b}{\Large\frac{m+60}{2}}$$\times 1000=60$-------(3)

From (1),(2) &(3)

$m=42$

$\therefore$ metal is Ca

Hence (b) is the correct answer.

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