Respuesta :
Answer: The amount of hydrogen gas collected will be 0.074 g
Explanation:
We are given:
Vapor pressure of water = 23.78 mmHg
Total vapor pressure = 742 mmHg
Vapor pressure of hydrogen gas = Total vapor pressure - Vapor pressure of water = (742 - 23.78) mmHg = 718.22 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 = 718.22 mmHg
V = Volume of the gas = 0.955 L
T = Temperature of the gas = [tex]25^oC=[25+273]K=298K[/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]718.22mmHg\times 0.955L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{718.22\times 0.955}{62.3637\times 298}=0.037mol[/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 = 0.037 moles
Molar mass of hydrogen gas = 2 g/mol
Putting values in above equation, we get:
[tex]0.037mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.037mol\times 2g/mol)=0.074g[/tex]
Hence, the amount of hydrogen gas collected will be 0.074 g
Mass of hydrogen collected is 0.074 g of hydrogen gas.
The equation of the reaction is;
2H(aq) + Zn(s) → H2(g) + Zn2+ (aq)
The total pressure of the gas = 742 mmHg
Since the gas was collected over water, the vapor pressure of water = 23.7 mmHg
Hence partial pressure of hydrogen gas in the mixture = 742 mmHg - 23.7 mmHg = 718.3 mmHg
From the ideal gas equation;
PV =nRT
P = 718.3 mmHg or 0.945 atm
V = 0.955 L
n = ?
R = 0.082 atmLmol-1K-1
T = 25 ∘C + 273 = 298 K
n = 0.945 atm × 0.955 L/0.082 atmLmol-1K-1 × 298 K
n = 0.902/24.44
n = 0.037 moles
Mass of hydrogen = number of moles × molar mass
Mass of hydrogen = 0.037 moles × 2 g/mol
Mass of hydrogen = 0.074 g of hydrogen
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