Question

A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the extent of $20\%$ by weight. $3.0g$ of the mineral on heating lost $1.1g$ of $C{O}_2$. What is the $\%$ by weight of the other metal?

• A. $70\%$
• B. $33\%$
• C. $67\%$
• D. $30\%$

Answer 1

The correct option is D $30\%$

$\text{Let, two equimolar carbonates are}AC{O}_{3}andBC{O}_3$
$WeightofmetalA=3\times 0.20=0.6gAC{O}_3+BC{O}_3\to AO+BO+2C{O}_2$
$\text{Total two moles}C{O}_3^{2-}\text{provide 2 moles of}C{O}_2$
or
$\text{one mole}C{O}_3^{2-}\text{provide 1 mole of}C{O}_2$
or
$60gC{O}_3^{2-}\text{provide}44gofC{O}_2$
$\text{Total mass of}C{O}_3^{2-}\text{required to produce 1.1 g of}C{O}_2=\frac{60}{44}\times 1.1=1.5g$
$weightofmetal(A+B)=3-1.5=1.5g$
$weightofB=1.5-0.6=0.9g$
$\%ofB=\frac{0.9}{3}\times 100=30\%$