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?

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

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation: