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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?

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Answer:

741 torr

0.81 moles of H2

Explanation:

Given the reaction equation;

Mg(s)+2HCl(aq)  →  MgCl2(aq)+H2(g)

We can obtain the pressure of the gas at 298 K using Dalton's law of partial pressures.

Pressure of gas + vapour pressure of water = 765 torr

vapour pressure of water = 24 torr

765 torr -  vapour pressure of water = Pressure of gas

Pressure of gas = 765 torr - 24 torr = 741 torr

From the question we have;

number of moles of HCl reacted = Concentration * volume

number of moles of HCl reacted = 250/1000 * 6.44 = 1.61 moles

If

2 moles of HCl yields 1 mole of H2

1.61 moles of HCl yields 1.61 * 1/2 = 0.81 moles of H2

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