Question

2.76 g of silver carbonate on being strongly heated yields a residue weighing
A ) 2.56 g
B ) 2.81 g
C ) 1.23 g
D ) 2.16 g

Answer 1

Hint: Thermal decomposition of one mole of silver carbonate gives two moles of silver. Use this information to write a balanced equation and hence use stoichiometry to calculate the answer.

 Complete step by step answer:
When silver carbonate is strongly heated, silver metal, carbon dioxide gas and oxygen gas are obtained. Carbon dioxide gas and oxygen gas escape in the atmosphere whereas residue of silver metal remains. Write the balanced chemical equation for the thermal decomposition of silver carbonate:
\[{\rm{A}}{{\rm{g}}_2}{\rm{C}}{{\rm{O}}_3}\left( s \right) \xrightarrow{\text{strong heating}} 2 Ag\left( s \right){\rm{ + C}}{{\rm{O}}_2}\left( g \right){\rm{ + }}\dfrac{1}{2}{{\rm{O}}_2}\left( g \right)\]

The atomic weights of silver, carbon and oxygen are 108 g/mol, 12 g/mol and 16 g/mol respectively. The molecular weight of silver carbonate is 2(108)+12+3(16)=276 g
The weight of silver carbonate is 2.76 g. Divide weight with molecular weight to obtain the number of moles of silver carbonate.
\[\dfrac{{2.76{\rm{ g}}}}{{276{\rm{ g/mol}}}} = 0.01{\rm{ mol}}\]
Use the reaction stoichiometry to calculate the number of moles of silver produced when 0.01 moles of silver carbonate are strongly heated.
\[{\rm{0}}{\rm{.01 mol A}}{{\rm{g}}_2}{\rm{C}}{{\rm{O}}_3}{\rm{ }} \times {\rm{ }}\dfrac{{{\rm{2 mol Ag}}}}{{1{\rm{ mol A}}{{\rm{g}}_2}{\rm{C}}{{\rm{O}}_3}}}{\rm{ = 0}}{\rm{.02 mol Ag}}\]

The atomic mass of silver is 108 g/mol. Calculate the mass of silver obtained.
\[{\rm{0}}{\rm{.02 mol }} \times {\rm{ 108 g/mol = 2}}{\rm{.16 g }}\]

Hence, option D ) is the correct answer.

Note:
Do not add mass of carbon dioxide or oxygen to the mass of silver obtained as carbon dioxide and oxygen are gases and will escape in the atmosphere.