
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
570.9k+ views
Hint: 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.
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.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

An example of chemosynthetic bacteria is A E coli B class 11 biology CBSE

10 examples of friction in our daily life

