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Question

Question: \[CaC{o_3}\] is \[90\% \] pure volume of \[C{O_2}\] collected at STP when \[10{\text{ }}gram\] of \[...

CaCo3CaC{o_3} is 90%90\% pure volume of CO2C{O_2} collected at STP when 10 gram10{\text{ }}gram of CaCO3CaC{O_3} is decomposed is:
A.  2.0 16 liters\;2.0{\text{ }}16{\text{ }}liters
B.1.008 liters1.008{\text{ }}liters
C.10.08 liters10.08{\text{ }}liters
D.  20.16 liters\;20.16{\text{ }}liters

Explanation

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 STPSTP and has values such as 00 degrees Celsius and 11 atmosphere of pressure.
According to question the reaction occurs is CaCO3 CaO+CO2CaC{O_3} \to {\text{ }}CaO + C{O_2}
From the reaction the gas produced is CO2C{O_2}
Therefore,
1{\text{ }}\;mol\;$$$$CaC{O_3} produces 1{\text{ }}\;mol$$$$C{O_2}
So, we have 90100 × 10 = 9g \dfrac{{90}}{{100}}{\text{ }} \times {\text{ }}10{\text{ }} = {\text{ }}9g{\text{ }} pure CaCO3CaC{O_3}
Molar mass of CaCO3=100g/molCaC{O_3} = 100g/mol
So, 9g CaCO3= 9100 = 0.09 mol CaCO39g{\text{ }}CaC{O_3} = {\text{ }}9100{\text{ }} = {\text{ }}0.09{\text{ }}mol{\text{ }}CaC{O_3}
This will produce 0.09   mol CO20.09{\text{ }}\;mol{\text{ }}C{O_2}
At STP 1mol = 22.4LSTP{\text{ }}1 mol{\text{ }} = {\text{ }}22.4L
0.09  mol = 0.09 x 22.4 = 2.016L0.09\;mol{\text{ }} = {\text{ }}0.09{\text{ }}x{\text{ }}22.4{\text{ }} = {\text{ }}2.016L
Hence, the volume of CO2C{O_2} collected at STP when 10 gram10{\text{ }}gram of CaCO3CaC{O_3} is decomposed is 2.0L2.0L

Hence the correct option is A.

Note: The standard temperature and pressure or we can say STPSTP is said to be a system which have a temperature value of zero degrees centigrade that is 273273 Kelvin and the pressure value is equal to the atmosphere will always be 11 atm. as well, one mole of any gas at STP value occupies a volume of about 22.414 Litre22.414{\text{ }}Litre. Though the concept of standard temperature and volume only takes place for the true gases only.