$4.08 ext{ g}$ of a mixture of $ ext{BaO}$ and an unknown alkaline earth metal carbonate $ ext{M

Question

$4.08 	ext{ g}$ of a mixture of $	ext{BaO}$ and an unknown alkaline earth metal carbonate $	ext{MCO}_3$ was heated strongly. The residue weighed $3.64 	ext{ g}$. This residue was dissolved in $100 	ext{ mL}$ of $1 	ext{ N HCl}$. The excess acid required $16 	ext{ mL}$ of $2.5 	ext{ N NaOH}$ solution for complete neutralization. What is the atomic weight of the metal $M$? (Given atomic weights: $	ext{Ba}=138$, $	ext{C}=12$, $	ext{O}=16$)

Professor PC Thomas Classes · Answer

Heating causes loss of $	ext{CO}_2$: $	ext{MCO}_3 ightarrow 	ext{MO} + 	ext{CO}_2$. 
Mass loss $= 4.08 - 3.64 = 0.44 	ext{ g}$ of $	ext{CO}_2$. 
Moles of $	ext{CO}_2 = 0.44 / 44 = 0.01 	ext{ mol}$. Thus, moles of $	ext{MCO}_3 = 0.01 	ext{ mol}$ and moles of $	ext{MO}$ in residue $= 0.01 	ext{ mol}$. 
Mass of $	ext{MCO}_3 = 0.01(M + 60) 	ext{ g}$. Mass of $	ext{BaO} = 4.08 - 0.01(M + 60) 	ext{ g}$. 
Equivalents of $	ext{HCl}$ taken $= 100 	imes 1 = 100 	ext{ meq} = 0.1 	ext{ eq}$. 
Equivalents of $	ext{NaOH}$ used $= 16 	imes 2.5 = 40 	ext{ meq} = 0.04 	ext{ eq}$. 
Equivalents of acid reacted with residue $= 0.1 - 0.04 = 0.06 	ext{ eq}$. 
Since both $	ext{BaO}$ and $	ext{MO}$ are divalent, $0.06 	ext{ eq}$ corresponds to $0.03 	ext{ mol}$ of total basic oxides. 
Moles of $	ext{BaO} = 0.03 - 0.01 = 0.02 	ext{ mol}$. 
Molar mass of $	ext{BaO} = 138 + 16 = 154 	ext{ g/mol}$. 
Mass of $	ext{BaO} = 0.02 	imes 154 = 3.08 	ext{ g}$. 
Mass of $	ext{MCO}_3 = 4.08 - 3.08 = 1.00 	ext{ g}$. 
$0.01(M + 60) = 1.00 \implies M + 60 = 100 \implies M = 40$.