You have a mixture that contains 0.380 moles of Ne(g), 0.250 moles of He(g), and 0.500 moles CH4(g) at 400 K and 7.25 atm. What
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Answer:
Molar mass of the metal = 59.0 g/mol
Explanation:
Vapor pressure of water at = 23.78 torr
We are given:
Total vapor pressure = 756.0 torr
Vapor pressure of hydrogen gas = Total vapor pressure - Vapor pressure of water = (756.0 - 23.78) torr = 732.22 torr
To calculate the amount of hydrogen gas collected, we use the equation given by ideal gas which follows:
where,
P = pressure of the gas = 732.22 torr
V = Volume of the gas = 255 mL = 0.255 L ( 1 mL = 0.001 L )
T = Temperature of the gas =
R = Gas constant =
n = number of moles of hydrogen gas = ?
Putting values in above equation, we get:
From the reaction shown below:-
1 mole of hydrogen gas is produced when 1 mole of the metal is reacted.
Also,
0.01 mole of hydrogen gas is produced when 0.01 mole of the metal is reacted.
Moles of the metal = 0.01 mol
Mass taken = 0.590 g
The formula for the calculation of moles is shown below:
Thus,
Molar mass of the metal = 59.0 g/mol