Question

0.92g of $A{{g}_{2}}C{{O}_{3}}$ is heated strongly beyond melting point. After heating, the amount of residue is:
(A) 0.36
(B) 0.39
(C) 0.77
(D) 0.72

Answer 1

Hint: Write the reaction happening during this process. Calculate the number of moles for both reactants and products at the start of the reaction and after heating is completed. To the number moles, multiply the molecular weight of the residue and find out the amount of residue left after heating.

Complete step by step answer:
-Silver carbonate, $A{{g}_{2}}C{{O}_{3}}$ on strong heating decomposes to form silver metal, Ag, oxygen gas, ${{O}_{2}}$ and carbon dioxide, $C{{O}_{2}}$ is released.
-It is given by the reaction, $A{{g}_{2}}C{{O}_{3}}\xrightarrow{\Delta }2Ag+{}^{1}/{}_{2}{{O}_{2}}+C{{O}_{2}}$
-0.92g of $A{{g}_{2}}C{{O}_{3}}$ was taken as sample size for heating. Molecular weight of $A{{g}_{2}}C{{O}_{3}}$ is approximately 276g/mol.
-\[Number\,of\,moles\,of\,A{{g}_{2}}C{{O}_{3}}=\dfrac{0.92}{276}=0.0033\]
-Molar mass of Ag =108g/mol
-Two moles of Ag are formed on decomposition of $A{{g}_{2}}C{{O}_{3}}$. So, 0.0066 moles of Ag are formed.

	\[A{{g}_{2}}C{{O}_{3}}\xrightarrow{\Delta }2Ag\]
Initially	0.0033	0
After decomposition	0	0.0066

-Now, we need to find the amount of residue left. Since oxygen and carbon dioxide are gases, they will escape into the atmosphere. Only silver metal residue will be left behind.
-So, multiply the molecular weight of silver metal with the number of moles of silver present after complete decomposition.
\[Amount\,of\,Ag=0.0066\times 108=0.718g\]
So, we obtain the weight left behind is 0.72g.
So, the correct answer is “Option D”.

Note: Melting point is the temperature at which a solid is transformed to liquid. Here, the starting compound is heated beyond melting point, which leads to decomposition. Read the question carefully to avoid confusion.