Question

A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the extent of $12.5\%$ by weight. $2.8g$ of the mineral on heating lost $1.32g$ of $C{O}_2$. What is the $\%$ by weight of the other metal?

• A. $87.5$
• B. $35.71$
• C. $65.11$
• D. $23.21$

Answer 1

The correct option is D $23.21$

Let the two equimolar carbonates be $AC{O}_3$ and $BC{O}_3$.
Weight of metal $A=2.8\times 0.125=0.35g$.
$AC{O}_3+BC{O}_3\to AO+BO+2C{O}_2$.
Total two moles $C{O}_3^{2-}$ provide 2 moles of $C{O}_2$ or $60gC{O}_3^{2-}$ provide $44$ g of $C{O}_2$.
Total mass of $C{O}_3^{2-}$ required to produce $1.32$ g of $C{O}_2=\frac{60}{44}\times 1.32=1.8$ g.
Weight of metal $(A+B)=2.8-1.8=1g$.
Weight of $B=1-0.35=0.65g$
$\%ofB=\frac{0.65}{2.80}\times 100=23.21$.