Explanation

(a)
$\ \ \ 2Ag_{2}CO_{3}\overset{\ \ \ \ \mathrm{\Delta}\ \ \ \ }{\rightarrow}4\ Ag + 2CO_{2} + O_{2}$
$2 \times 276$ g $4 \times 108$ g ($s$)
$\because 2 \times 276$ g of $Ag_{2}CO_{3}$ gives=$4 \times 108$ g Ag
$\therefore$ 1 g of $Ag_{2}CO_{3}\$ gives$= \frac{4 \times 108}{2 \times 276}$
$\therefore$ 276 g of $Ag_{2}CO_{3}$ gives$= \frac{4 \times 108 \times 2.76}{2 \times 276}$
=2.16 g