Question

Volume of $C{O_2}$ obtained at STP by the complete decomposition of 9.85gm $BaC{O_3}$is:
A. $2.24\,\,litre$
B. $1.12\,litre$
C. $0.85\,litre$
D. $0.56\,litre$

Answer 1

The amounts of gas are compared at a set of standard conditions of temperature and pressure, abbreviated as STP. At STP, one mole of any gas has the same volume ${\text{22}}{\text{.4 L}}$ called the standard molar volume.
Example:
Under STP conditions, one mole of nitrogen gas and one mole of helium gas each contain ${\text{6}}{{.02 \times 1}}{{\text{0}}^{{\text{23}}}}{\text{ }}$ particles of gas and occupy a volume of ${\text{22}}{\text{.4 L}}$at ${{\text{0}}^{\text{o}}}{\text{C}}$ and ${\text{1atm}}$ pressure. Since the molar masses of nitrogen and helium are different (${\text{28}}{\text{.02 g}}$ for ${{\text{N}}_{\text{2}}}$ compared to ${\text{4}}{\text{.003 g}}$ for ${\text{He}}$) one mole of each substance has a different mass.

Complete answer: or Complete step by step answer:
Given data contains
The mass of barium carbonate as $9.85\,gm$.
The molecular weight of barium carbonate is $197\,g/mol$.
We know that volume of one mole of carbon dioxide is ${\text{22}}{\text{.4 L}}$
Volume of carbon dioxide obtained from complete decomposition of 9.85 gm of barium carbonate is:
We can give the decomposition reaction of barium carbonate as,
$BaC{O_3} \to BaO + C{O_2}$
We have to calculate the moles of barium carbonate using the given mass and molar mass.
The grams can be converted to moles as,
$grams \times \dfrac{{1\,mol}}{{molar\,mass}}$
$9.85{\text{ }}g \times \dfrac{{1\,mol}}{{197g}} = 0.05\,mol$
The calculated moles of barium carbonate is $0.05\,mol$.
From the chemical equation, we can observe that one mole of barium carbonate produces one mole of carbon dioxide.
Therefore, the number of moles of carbon dioxide produced is $0.05\,mol$.
The moles can be converted to liters by,
$n \times \dfrac{{22.4L}}{{1\,mol}}$
$0.05\,mol \times \dfrac{{22.4L}}{{1\,mol}} = 1.12\,L$
The volume of carbon dioxide obtained at STP is $1.12\,litre$.

So, the correct answer is Option B .

Note:
We must remember that the equal volume of all gases at the standard temperature and pressure comprises the same number of particles. Molar volume for crystalline solids could be measured with the help X-ray crystallography. For ideal gases, the ideal gas equation gives the molar volume.
Let us calculate the molar volume for the given example.
Example 1: Calculate the molar volume for a sample of the molar mass of nitrogen if the density of the gas is $1.270\,g/L.$
In this case, we can calculate the molar volume of the sample using the molar mass and density.
We know the molar mass of nitrogen is $28.01\,g/mol$.
The given density is $1.270\,g/L.$
We can calculate the molar volume by dividing the molar mass to the density. We can write the formula to calculate the molar volume as,
Molar volume=$\dfrac{{Molar\,mass}}{{Density}}$
Molar volume=$\dfrac{{28.01\,g/mol}}{{1.270\,g/L}}$
Molar volume=$22.05\,L$
The molar volume of gas sample is $22.05\,L$.