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

Last updated date: 09th Aug 2024
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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.

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 $2.16g$.