Solveeit Logo

Question

Question: Volume of \(C{{O}_ {2}} \) obtained at STP by the complete decomposition of 9.85gm \(BaC{{O}_ {3}} \...

Volume of CO2C{{O}_ {2}} obtained at STP by the complete decomposition of 9.85gm BaCO3BaC{{O}_ {3}} is:
(molecular weight of BaCO3BaC{{O}_ {3}} =197)
(A) 2.24 litre
(B) 1.12 litre
(C) 0.85 litre
(D) 0.56 litre

Explanation

Solution

A decomposition reaction is complete when the mass of the container and its contents no longer changes on heating. Continue the cycle until there is no further loss of mass on heating. At this point, you know that the decomposition is complete.

Complete step by step solution:
We have been given that BaCO3BaC{{O}_ {3}} is decomposed:
BaCO3BaO+CO2BaC{{O}_ {3}} \to BaO+C{{O}_ {2}}
We need to find the volume of CO2C{{O}_ {2}} obtained at STP by the complete decomposition of 9.85gm BaCO3BaC{{O}_ {3}} ,
So, for that firstly,
We have been given that CO2C{{O}_ {2}} is obtained at STP,
Standard temperature and pressure are standard sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data.
The volume of 1 mole of CO2C{{O}_ {2}} at STP= 22.4 litre,
We have been given the molecular weight of BaCO3BaC{{O}_ {3}} =197,
Now, we will be finding the moles of BaCO3BaC{{O}_ {3}} obtained: 9.85197=0.05mole\dfrac {9.85}{197} =0.05mole,
As 1 mole of BaCO3BaC{{O}_ {3}} gives 1 mole of CO2C{{O}_ {2}} ,
Therefore, number of moles of CO2C{{O}_ {2}} obtained= 0.05 mole,
So, the volume of CO2C{{O}_ {2}} obtained at STP: 22.4×0.05=1.12litre22.4\times 0.05=1.12litre
So, we can say that Volume of CO2C{{O}_ {2}} obtained at STP by the complete decomposition of 9.85gm BaCO3BaC{{O}_ {3}} is 1.12 litre.

Therefore, we can conclude that option (A) is correct.

Note: Decomposition, the opposite of combination, occurs when a single substance breaks down into two or more different substances. This type of chemical change is evident when fruit decomposes over time.