Respuesta :
Answer:
The amount of hydrogen gas in the sample is 0.0408 g
Explanation:
We are given:
Vapor pressure of water = 18.7 mmHg
Total vapor pressure = 748 mmHg
Vapor pressure of hydrogen gas = Total vapor pressure - Vapor pressure of water = (748 - 18.7) mmHg = 729.3 mmHg
To calculate the amount of hydrogen gas collected, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 729.3 mmHg
V = Volume of the gas = 512 mL = 0.512 L ( 1 mL = 0.001 L)
T = Temperature of the gas = [tex]21^oC=[21+273]K=294\ K[/tex]
R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
n = number of moles of hydrogen gas = ?
Putting values in above equation, we get:
[tex]729.3mmHg\times 0.512L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 294K\\\\n=\frac{729.3\times 0.512}{62.3637\times 294}=0.0204mol[/tex]
To calculate the mass from given number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of hydrogen gas = 0.0204 moles
Molar mass of hydrogen gas = 2 g/mol
Putting values in above equation, we get:
[tex]0.0204mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.0204mol\times 2g/mol)=0.0408g[/tex]
Hence, the amount of hydrogen gas in the sample is 0.0408 g
The grams of H2 are in the sample is mathematically given as
Mas of H2=0.0408g
What are the grams of H2 are in the sample?
Question Parameter(s):
collected over water at 21°C and
at a pressure equal to the atmospheric pressure of 748 mm Hg.
The volume of the sample was 512 mL.
Generally, the equation for the ideal gas is mathematically given as
PV=nRT
Therefore
[tex]n=\frac{729.3* 0.512}{62.3637* 294}[/tex]
n=0.0204mol
In conclusion
[tex]n=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Hence
Mass of H=0.0204 *2
Mas of H=0.0408g
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