Question
Question: \[CaC{o_3}\] is \[90\% \] pure volume of \[C{O_2}\] collected at STP when \[10{\text{ }}gram\] of \[...
CaCo3 is 90% pure volume of CO2 collected at STP when 10 gram of CaCO3 is decomposed is:
A.2.0 16 liters
B.1.008 liters
C.10.08 liters
D.20.16 liters
Solution
In the given question we have to find out the decomposed calcium carbonate according to the value of standard temperature and volume. Initially we will have to solve by using the concept of moles. After finding the moles of each we can easily find out the volume.
Complete Step by step answer: The ideal gas law in the chemistry states that the volume engaged by a gas depends upon the amount of substance (which is gas) as well as temperature and on pressure. And the value of Standard temperature and standard pressure is usually abbreviated by the term STP and has values such as 0 degrees Celsius and 1 atmosphere of pressure.
According to question the reaction occurs is CaCO3→ CaO+CO2
From the reaction the gas produced is CO2
Therefore,
1{\text{ }}\;mol\;$$$$CaC{O_3} produces 1{\text{ }}\;mol$$$$C{O_2}
So, we have 10090 × 10 = 9g pure CaCO3
Molar mass of CaCO3=100g/mol
So, 9g CaCO3= 9100 = 0.09 mol CaCO3
This will produce 0.09 mol CO2
At STP 1mol = 22.4L
0.09mol = 0.09 x 22.4 = 2.016L
Hence, the volume of CO2 collected at STP when 10 gram of CaCO3 is decomposed is 2.0L
Hence the correct option is A.
Note: The standard temperature and pressure or we can say STP is said to be a system which have a temperature value of zero degrees centigrade that is 273 Kelvin and the pressure value is equal to the atmosphere will always be 1 atm. as well, one mole of any gas at STP value occupies a volume of about 22.414 Litre. Though the concept of standard temperature and volume only takes place for the true gases only.