Question

Calculate the residue obtained on strongly heating $2.76g$ $A{g_2}C{O_3}$?
$A{g_2}C{O_3}\xrightarrow{\Delta }2Ag + C{O_2} + \dfrac{1}{2}{O_2}$

Answer 1

Hint: As we know that one mole is defined as the ratio of given mass of a particular substance to the molecular mass of that substance which is again equivalent to the ratio of number of molecules or atoms to the Avogadro’s number. We also know that one mole of any reactant produces the equivalent moles of the product.

Complete answer:
We know that mole is the definite amount of a substance which can be expressed either in terms of weight of that substance, volume of the substance as well as number of particles of that particular substance.
So, as we can see from the above given equation that one mole of silver carbonate decomposes to form two moles of silver, one mole of carbon dioxide and half mole of oxygen. In other words we can say that $275.7g$ of silver carbonate decomposes to form $216g$ of silver and $44g$ of carbon dioxide.
Now we are given that $2.76g$ of silver carbonate is decomposing.
So, if $275.7g$of silver carbonate forms $216g$ of silver then, $2.76g$ of silver carbonate will form:
$ \Rightarrow \dfrac{{216 \times 2.76}}{{276}} = 2.16g$

Therefore, the residue which is obtained on strong heating of silver carbonate is $2.16g$.

Note: Always remember that mole concept is the basics for finding any parameter like moles of any substance, mass of a substance, volume and number of atoms and molecules present in a compound because one mole contains a fixed number of particles. We can also calculate molarity, normality and molality using the mole concept. Also remember the stoichiometric coefficients of the reactants as well products for solving any parameter.