Question

What is the weight of the residue obtained when \[2.76g\] of silver carbonate is strongly heated.
A.\[2.16g\]
B.\[2.48g\]
C.\[2.32g\]
D.\[2.64g\]

Answer 1

Hint: Strongly heating silver carbonate results in a decomposition reaction. The residue obtained by a decomposition reaction will be the solid product or the solid remains, the gaseous products tend to escape and are not considered as a residue.

Complete answer:
A decomposition reaction is a type of chemical reaction in which a single reactant cleaves or breaks down into its fragments by the action of heat or sunlight on it. This gives more than one product from a single reactant.
Metal carbonates when heated strongly have a tendency to get decomposed and result in the evolution of carbon dioxide gas. Silver carbonate also undergoes a similar reaction upon getting strongly heated.
Chemical reactions follow certain rules known as the rules of chemical combination. One such rule is the law of conservation of mass according to which creating or destroying matter is not possible. Hence the total mass remains conserved in a reaction. When we talk in terms of moles, we can say that only a specific number of moles of the reactant react in a definite ration to produce products in definite ratios. Thus the stoichiometric numbers are used to indicate the ratios of moles in which reactants react to produce products so as to keep the total mass conserved.
The decomposition reaction of silver carbonate can be written as follows:
\[{\text{A}}{{\text{g}}_2}C{O_3}\xrightarrow{{heating}}2Ag(s) + C{O_2}(g) + \dfrac{1}{2}{O_2}(g)\]
Thus one mole of silver carbonate gives two moles of silver metal residue on decomposition. The carbon dioxide and oxygen gas escapes on heating. Therefore, we know that silver carbonate and silver metal are reacted formed in the ratio \[1:2\] .
The number of moles present in \[2.76g\] of silver carbonate can be calculated as follows:
\[{\text{number of moles}} = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}\]
Inserting the values of given mass and molar mass,
\[{\text{given mass}} = 2.76g\]
\[{\text{molar mass of A}}{{\text{g}}_2}C{O_3} \approx 276gmo{l^{ - 1}}\]
\[{\text{number of moles of A}}{{\text{g}}_2}C{O_3} = \dfrac{{2.76g}}{{276gmo{l^{ - 1}}}}\]
The number of moles of silver carbonate are,
\[{\text{number of moles of A}}{{\text{g}}_2}C{O_3} = 0.01moles\]
The moles of silver metal produced will be twice the number of reacting moles of silver carbonate. The mass of silver metal residue can be calculates as follows:
\[{\text{mass of Ag(s) = (number of moles)}} \times {\text{molar mass}}\]
Inserting the molar mass of silver in the above equation:
\[{\text{molar mass of Ag}} \approx 108gmo{l^{ - 1}}\]
\[{\text{mass of Ag(s) = (2}} \times 0.01moles{\text{)}} \times 108gmo{l^{ - 1}}\]
\[{\text{mass of Ag(s) }} = 2.16g\]
Hence, the correct option is (A) and the mass of residue obtained is \[{\text{mass of Ag(s) }} = 2.16g\] .

Note:
Silver is a transition metal and can therefore show variable oxidation state. Silver has the tendency to show the oxidation number one or two. The silver carbonate that gets decomposed is the one containing silver with \[ + 1\] oxidation state and hence the stoichiometry should be decided accordingly.