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4.08g of a mixture of BaO and unknown carbonate $(MCO_3)$ was heated strongly.The residue of 3.64g was dissolved in 100ml of 1N HCl.The excess acid required 16ml of 2.5N NaOH solution for complete neutralization.Identify metal


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$BaO \quad\underrightarrow {\Delta}\quad$ No effect
$MCO_3 \quad\underrightarrow {\Delta}\quad MO+CO_2$
Loss in weight=4.08-3.64=0.44g
Weight of $CO_2=0.44g$
Also weight of $CO_2$ given by 'b' g $MCO_3$
$\Rightarrow \large\frac{44b}{m+60}$
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)
$\therefore$ metal is Ca
Hence (b) is the correct answer.
answered Jan 7, 2014 by sreemathi.v

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