Question

The weight of residue obtained on strongly heating $1.38g$ $A{g_2}C{O_3}$ is:
A.$2.16g$
B.$1.08g$
C.$1.16g$
D.$232g$

Answer 1

Hint:The term ‘residue’ given in question is defined as the left part or the remaining part obtained in product after any chemical process of preparation. $A{g_2}C{O_3}$ (Silver (I) carbonate) on heating strongly undergoes thermal decomposition. The best approach towards this type of problems is to write a balanced chemical reaction in terms of moles (ratio of given weight to the molecular weight).

Complete step by step answer:
First step is to write a balanced chemical reaction. $A{g_2}C{O_3}$ Strongly heating undergoes thermal decomposition to produce silver, carbon dioxide and oxygen.
$2A{g_2}C{O_3}\xrightarrow{\Delta }4Ag + 2C{O_2} \uparrow + {O_2} \uparrow $
According to the above reaction residue obtained is $Ag$. We can easily understand that two $2$ moles of $A{g_2}C{O_3}$ on heating produces $4$ moles of $Ag$ or simply $1$ mole of $A{g_2}C{O_3}$ produces $2$ moles of $Ag$.
$2A{g_2}C{O_3}\xrightarrow{\Delta }4Ag + 2C{O_2} \uparrow + {O_2} \uparrow $
Now the weight of $A{g_2}C{O_3}$ is $1.38g$. To find number of moles of $A{g_2}C{O_3}$ we need molecular weight of $A{g_2}C{O_3}$ which can be calculated as follows:
Molecular weight $A{g_2}C{O_3}$$ = 108 \times 2 + 12 + 16 \times 3 = 276g$, $\left( {Ag = 108,C = 12,O = 16} \right)$
We know that the number of moles is the ratio of given weight to the molecular weight and it is denoted by n. Let given weight be w and molecular weight be M. We have $w = 1.38g, M = 276g$. Now substitute the values we get,
$n = {w}{M} = {{1.38}}{{276}} = 0.005$ Moles
Now again consider the chemical reaction,
$2A{g_2}C{O_3}\xrightarrow{\Delta }4Ag + 2C{O_2} \uparrow + {O_2} \uparrow $
From the above reaction we can conclude that $1$ mole of $A{g_2}C{O_3}$ produces $2$ moles of $Ag$.
Therefore, $0.005$ moles of $A{g_2}C{O_3}$ will produce $0.005 \times 2 = 0.01$ moles of residue $Ag$.
We calculated that $0.01$ moles of residue will be obtained on strongly heating $1.38g$ $A{g_2}C{O_3}$ . But we have options in grams so now we will convert $0.01$ moles in grams using the relation,
$w = n \times M$ , substitute the values in given formula to get the desired result,
$w = 0.01 \times 108 = 1.08g$ $\left( {{M_{Ag}} = 108g} \right)$
Final Result: : The weight of residue obtained on strongly heating $1.38g$$A{g_2}C{O_3}$ is $1.08g$

Thus, the correct option is (B).


Note:
-‘Residue’ is also known as a by-product of chemical reaction.
-While comparing moles of chemical compounds in reactants with products consider the smallest ratio between them. For example $2:4$ must be considered as $1:2$.
-$A{g_2}C{O_3}$ (Silver (I) carbonate) is yellow but some samples are grayish due to the presence of element silver.
-$A{g_2}C{O_3}$ (Silver (I) carbonate) is a transition metal carbonate and it is poorly soluble in water.