Question

Amount of $N{a}_{2}C{O}_3$ reacted in millimoles is:

• A. $1.83\times {10}^{-4}$
• B. $1.7\times {10}^{-10}$
• C. $3.1\times {10}^{-4}$
• D. $2.1\times {10}^{-10}$

Answer 1

The correct option is A $1.83\times {10}^{-4}$

$N{a}_{2}C{O}_3+2AgCl\to A{g}_{2}C{O}_3+2NaCl$

Initial millimoles of $N{a}_{2}C{O}_3,AgCl,A{g}_{2}C{O}_3$ and $NaCl$ present are $(5\times 1.5=7.5)$ excess, 0 and 0 respectively.

After the reaction, millimoles of $N{a}_{2}C{O}_3,AgCl,A{g}_{2}C{O}_3$ and $NaCl$ remaining are $(7.5-a)$ excess, $a$ and $2a$ respectively.

The chloride ion concentration is $[C{l}^{-}]=\frac{0.0026}{35.5}=7.32\times {10}^{-5}$. It is equal to $=\frac{millimoles}{VmL}=\frac{2a}{5}$

Hence, $,\frac{2a}{5}=7.32\times {10}^{-5}$ or amount of sodium carbonate reacted is $a=1.83\times {10}^{-4}millimoles$.