Question
Question: Calculate the mass of residue obtained on strongly heating 2.76g of \(A{{g}_{2}}C{{O}_{3}}\). \(A{...
Calculate the mass of residue obtained on strongly heating 2.76g of Ag2CO3.
Ag2CO3Δ2Ag+CO2+21O2
Solution
Approach this question by using mole concept which will help us to find the moles of silver carbonate and this will be further useful in calculation of moles of the residue on the basis of the given reaction. The reaction must be balanced.
Formula used:
We will use the following formulas in this solution:-
n=molar mass mass
where n is the number of moles of a substance.
Complete answer:
Let us first discuss about the silver carbonate as follows:-
Silver carbonate: It is a metallic carbonate whose chemical formula is Ag2CO3.It is quite basic in nature and when it is heated at high temperature, it produces silver element along with the liberation of carbon dioxide and oxygen gas.
-Calculation of moles of silver carbonate ( Ag2CO3):-
Atomic mass of silver = 108 g/mol
Atomic mass of carbon = 12 g/mol
Atomic mass of oxygen = 16 g/mol
Molar mass of Ag2CO3= 2(108 g/mol) + 12 g/mol +3(16 g/mol) =276g/mol
Given mass of Ag2CO3= 2.76g
⇒n=molar mass mass⇒n=276g/mol2.76g⇒n=0.01moles
-The strong heating of Ag2CO3gives the following reaction:-
Ag2CO3Δ2Ag+CO2+21O2
As we can see that the residue produced is a silver element. It is clear that 1 mole of silver carbonate produced 2 moles of silver. So 0.01 mole of silver carbonate will produce 0.02 moles of silver.
-Calculation of mass of silver obtained as follows:-
n=molar mass mass
Rearrange the formula for further calculations:-
⇒ mass=n×molar mass⇒ mass=0.02mole×108g/mol⇒ mass=2.16g
-The mass of residue (Ag) obtained is 2.16grams.
Note:
-Always remember to balance the reaction (if not balanced in the question) before solving such types of questions because it would give correct information regarding the amount or moles of substance produced, combusted, oxidized or reduced in a reaction.