Question

The mass of residue left after strongly heating 1.38 g of silver carbonate will be:

• A. 1.16 g
• B. 1.33 g
• C. 2.66 g
• D. 1.08 g

Answer 1

The correct option is A 1.16 g

$A{g}_{2}C{O}_3\to A{g}_{2}O+C{O}_2\left(g\right)$
I mole of $A{g}_{2}C{O}_3$ (275.74) grams on decomposition gives a mass residue of 231.73 grams of $A{g}_{2}O$ remaning mass is liberated as $C{O}_2$ gas.
If 1.38 grams $A{g}_{2}C{O}_3$ is hetaed it produces X grams of mass residue of $A{g}_{2}O$
$X=\frac{1.38\times 231.739}{275.745}=1.16$ grams
Hence option A is correct.