Question

$4.08\, g$ of a mixture of $BaO$ and unknown carbonate $MCO _{3}$ was heated strongly. The residue weighed $3.64\, g$. This was dissolved in $100\, mL$ of $1\, N\, HCl$. The excess acid required $16 \,mL$ of $2.5 \,N \,NaOH$ solution for complete neutralisation. Identify the metal $M$.

Answer 1

Weight of mixture of $BaO$ and $M C O_{3}=4.08\, g$
On heating of $M C O_{3}$ gives $MO$ and $C O_{2}$
Weight of $B a O+M O=3.64 \, g$
Weight of $CO _{2}$ evolved $=4.08-3.64=0.44 \, g$
Mols of $CO _{2}=\frac{0.44}{44}=10^{-2}\, mol =10^{-2} \times 10^{3}$
$=10\, m\, mol =20 \, m$ Eq of $CO _{2}$
$=20 \, m\, Eq\, MCO _{3}$
Total acid $=100 \times 1=100\, m\, Eq$
Excess of acid $=$ Total acid $- m"$ Eq of " $N a O H$
$=100-16 \times 2.5=60 \,m\,Eq$ of excess acid
$\left(E w(B a O)=\frac{M w}{2}\right)$
$m$ Eq of $B a O=60-20=40 \, m\, Eq$
$=\frac{(138+16) \times 40}{100 \times 2}=3.08 \, g$
weight of $M O=3.64-3.08=0.56 \, g$
$20 m " Eq$ of " $M O=0.56$
Equivalent weight of $M O=0.56 \times 1000=28$
Molecular weight of $M O=28 \times 2=56$
Hence, atomic weight of $M=56-16=40$.