Question

2.76 g of silver carbonate on being strongly heated yields a residue weighing:
A) 3.54 g
B) 3.0 g
C) 1.36 g
D) 2.16 g

Answer 1

The chemical formula for silver carbonate is ${\text{A}}{{\text{g}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$. On strong heating it converts to a residue of silver metal and gives out carbon dioxide gas and oxygen gas.

Complete step by step answer:
Write the balanced chemical equation for the reaction when silver carbonate is heated as follows:
The chemical formula for silver carbonate is ${\text{A}}{{\text{g}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$. On strong heating it converts to a residue of silver metal and gives out carbon dioxide gas and oxygen gas . The balanced chemical equation for the reaction when silver carbonate is strongly heated is as follows:
{\text{2A}}{{\text{g}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\xrightarrow{{\text{\Delta }}}{\text{4Ag}} + {\text{2C}}{{\text{O}}_{\text{2}}} + {{\text{O}}_{\text{2}}}
Write the number of moles of silver carbonate and the number of moles of silver residue from the reaction stoichiometry as follows:
From the reaction stoichiometry, we can say that,
$2{\text{ mol A}}{{\text{g}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} = 4{\text{ mol Ag}}$
Calculate the mass of silver carbonate and silver residue from the reaction stoichiometry as follows:
The mass of any substance present in one mole of it is equal to its molar mass. Thus,
$2{\text{ mol A}}{{\text{g}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} = 2 \times 276{\text{ g A}}{{\text{g}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} = 552{\text{ g A}}{{\text{g}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ and
$4{\text{ mol Ag}} = {\text{4}} \times 108{\text{ g Ag}} = 432{\text{ g Ag}}$
Thus, $552{\text{ g}}$ of ${\text{A}}{{\text{g}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ gives a residue of $432{\text{ g}}$ of ${\text{Ag}}$ on strong heating.
Calculate the mass of residue when $2.76{\text{ g}}$ of silver carbonate is heated as follows:
We have $2.76{\text{ g}}$ of silver carbonate. Thus, the residue obtained on heating $2.76{\text{ g}}$ of ${\text{A}}{{\text{g}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is,
${\text{Mass of residue}} = 2.76{\text{ g A}}{{\text{g}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} \times \dfrac{{432{\text{ g Ag}}}}{{552{\text{ g A}}{{\text{g}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}$
${\text{Mass of residue}} = 2.16{\text{ g}}$
Thus, $2.76{\text{ g}}$ of silver carbonate on being strongly heated yields a residue weighing $2.16{\text{ g}}$.

Thus, the correct option is (D) $2.16{\text{ g}}$.

Note: The most important step is to write the correct balanced chemical equation. If the balanced chemical equation is wrong it leads to wrong stoichiometry and thus, leads to incorrect final answer.