Mg(s)+2HCl(aq) → MgCl2(aq)+H2(g)
In an experiment, a student places a small piece of pure Mg(s) into a beaker containing 250.mL of 6.44MHCl(aq) . A reaction occurs, as represented by the equation above.
The student collects the H2(g) produced by the reaction and measures its volume over water at 298 K after carefully equalizing the water levels inside and outside the gas-collection tube. The volume is measured to be 45.6mL . The atmospheric pressure in the lab is measured as 765 torr , and the equilibrium vapor pressure of water at 298 K is 24 torr .
Calculate the following.
(i) The pressure inside the tube due to the H2(g)
(ii) The number of moles of H2(g) produced in the reaction
what is the answer and why?
Answers
Answered by
30
The pressure inside the tube due to H2 is 741 torr
Explanation:
- By applying Dalton's law we can easily calculate it.
- The given total pressure in the tube is 765 torr. Also hydrogen gas and its accumulation over water contribute to the total pressure . So the pressure of water vapours is 24 torr.
- Using the equation P(total) = Pgas1 + Pgas2 + Pgas3, P(total) = P of hydrogen + P of water replacing this with values we get 765 torr = P + 24 torr and we get 741 torr.
- While Number of moles can be found out using the given formula = Mass of hydrogen / Mass of one mole of hydrogen so the answer is 1 mole of hydrogen is produced.
Answered by
15
The pressure inside the tube due to the H2(g) = 741 torr
1 mole of hydrogen is produced
Explanation:
(i) Atmospheric pressure in the lab = total pressure on the tube = 765 torr.
The hydrogen gas accumulation over water is included in total pressure.
Pressure of water vapour = 24 torr
By applying Dalton's Law, we know that:
P(total) = P1 + P2 + P3
Here P(total) = P of hydrogen + P of water
765 torr = P + 24 torr
Therefore Pressure of hydrogen = 765 - 24 = 741 torr.
(ii) The number of moles of H2(g) produced in the reaction = Mass of hydrogen / Mass of one mole of hydrogen = 1
So we find that 1 mole of hydrogen is produced.
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