
How do you calculate the volume of ${\text{25}}{\text{.0}}$ g carbon monoxide at STP?
Answer
558.6k+ views
Hint: To answer this question we should know the ideal gas equation and mole formula. First we will determine the moles of carbon dioxide by using the mole formula. Then by using the ideal gas equation and STP condition we can determine the volume of carbon dioxide.
Formula used: ${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$
${\text{pV = nRT}}$
Complete step-by-step answer:We will determine the mole of carbon monoxide as follows:
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$
The molar mass of carbon monoxide is $28.01$g/mol.
On substituting $25.0$g for moles and $28.01$g/mol for molar mass of carbon monoxide,
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{25}}{\text{.0}}}}{{28.01}}$
$\Rightarrow {\text{mole}}\,{\text{ = }}\,0.89$
So, the moles of carbon monoxide is $0.89$ mol.
The value of standard temperature in kelvin is $273\,{\text{K}}$. The value of the standard pressure in atm is${\text{1}}\,{\text{atm}}$.
The formula of the ideal gas is as follows:
${\text{pV = nRT}}$
${\text{p}}$is the pressure
V is the volume
${\text{n}}$ is the number of moles of ideal gas
R is the gas constant
T is the temperature
On rearranging the ideal gas equation for V we get,
${\text{V}}\, = \,\dfrac{{{\text{nRT}}}}{{\text{p}}}$
The value of R in atm is ${\text{0}}{\text{.0821}}\,{\text{Latm}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ .
On substituting ${\text{1}}\,{\text{atm}}$ for p, $0.89$ L for n, ${\text{0}}{\text{.0821}}\,{\text{Latm}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ for R and $273\,{\text{K}}$ for T,
$\Rightarrow {\text{V}}\,{\text{ = }}\dfrac{{{\text{0}}{\text{.89}}\,{\text{mol}} \times {\text{0}}{\text{.0821}}\,{\text{Latm}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}} \times 273\,{\text{K}}}}{{{\text{1}}\,{\text{atm}}}}$
$\Rightarrow {\text{V}}\,{\text{ = }}20.00\,{\text{L}}$
So, the volume of ${\text{25}}{\text{.0}}$ g carbon monoxide at STP is $20.00$ L.
Therefore, $20.00$ L is the correct answer.
Note: The STP is known as standard temperature and pressure. One mole of an ideal gas at $273\,{\text{K}}$ temperature and ${\text{1}}\,{\text{atm}}$ pressure occupies $22.4\,{\text{L}}$volume. Volume is directly proportional to the number of moles so, here, the volume is less than $22.4\,{\text{L}}$ because the number of moles is less than one. Molar mass of a molecule is determined by adding the atomic mass of each atom of the molecule. The unit of pressure and volume is important. The value of R changes in different units.
Formula used: ${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$
${\text{pV = nRT}}$
Complete step-by-step answer:We will determine the mole of carbon monoxide as follows:
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$
The molar mass of carbon monoxide is $28.01$g/mol.
On substituting $25.0$g for moles and $28.01$g/mol for molar mass of carbon monoxide,
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{25}}{\text{.0}}}}{{28.01}}$
$\Rightarrow {\text{mole}}\,{\text{ = }}\,0.89$
So, the moles of carbon monoxide is $0.89$ mol.
The value of standard temperature in kelvin is $273\,{\text{K}}$. The value of the standard pressure in atm is${\text{1}}\,{\text{atm}}$.
The formula of the ideal gas is as follows:
${\text{pV = nRT}}$
${\text{p}}$is the pressure
V is the volume
${\text{n}}$ is the number of moles of ideal gas
R is the gas constant
T is the temperature
On rearranging the ideal gas equation for V we get,
${\text{V}}\, = \,\dfrac{{{\text{nRT}}}}{{\text{p}}}$
The value of R in atm is ${\text{0}}{\text{.0821}}\,{\text{Latm}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ .
On substituting ${\text{1}}\,{\text{atm}}$ for p, $0.89$ L for n, ${\text{0}}{\text{.0821}}\,{\text{Latm}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ for R and $273\,{\text{K}}$ for T,
$\Rightarrow {\text{V}}\,{\text{ = }}\dfrac{{{\text{0}}{\text{.89}}\,{\text{mol}} \times {\text{0}}{\text{.0821}}\,{\text{Latm}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}} \times 273\,{\text{K}}}}{{{\text{1}}\,{\text{atm}}}}$
$\Rightarrow {\text{V}}\,{\text{ = }}20.00\,{\text{L}}$
So, the volume of ${\text{25}}{\text{.0}}$ g carbon monoxide at STP is $20.00$ L.
Therefore, $20.00$ L is the correct answer.
Note: The STP is known as standard temperature and pressure. One mole of an ideal gas at $273\,{\text{K}}$ temperature and ${\text{1}}\,{\text{atm}}$ pressure occupies $22.4\,{\text{L}}$volume. Volume is directly proportional to the number of moles so, here, the volume is less than $22.4\,{\text{L}}$ because the number of moles is less than one. Molar mass of a molecule is determined by adding the atomic mass of each atom of the molecule. The unit of pressure and volume is important. The value of R changes in different units.
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