A student performs an experiment to determine the volume of hydrogen gas produced when a given mass of magnesium reacts with excess HCl(aq), as represented by the net ionic equation above. The student begins with a 0.0360 g sample of pure magnesium and a solution of 2.0 M HCl(aq). (a) Calculate the number of moles of magnesium in the 0.0360 g sample. b) Calculate the number of moles of HCl(aą) needed to react completely with the sample of magnesium. As the magnesium reacts, the hydrogen gas produced is collected by water displacement at 23.0°'C. The pressure of the gas in the collection tube is measured to be 749 torr (c) Given that the equilibrium vapor pressure of water is 21 torr at 23.0°C, calculate the pressure produced in the reaction would have if it were dry. (d) Calculate the volume, in liters measured at the conditions in the laboratory, that the Hy(e) produced in the reaction would have if it were dry (e) The laboratory procedure specified that the concentration of the HCl solution be 2.0 M, but only 12.3 M HCl solution was available. Describe the steps for safely preparing 50.0 mL of 2.0 M HCKag) using 12.3 M HCI solution and materials selected from the list below. Show any necessary calculation(s). Distilled water 10.0 mL graduated cylinder Balance 250 mL beakers Dropper 50.00 mL volumetric flask

(c) Since we are collecting the Hydrogen gas produced in the reaction over water we need to substract the water vapor pressure from the pressure measured in the lab to obtain the dry pressure:

Pdry = 749 torr - 21 torr = 728 torr

(d) The volume of the Hydrogen gas is obtained from the ideal gas law since we know the temperature and the dry pressure:

PV = nRT ∴ V = nRT/ P

we would need first to convert the pressure to atmospheres:

P= 728 torr x 1 atm/760 torr = 0.96 atm

Then,

mol H₂ gas produced:

From the balanced chemical equation,

1 mol H2/ 1 mol Mg x 0.015 mol Mg = 0.0015 mol

Now we have all we need to calculate the volume:

V = 0.0015 mol x 0.0821 Latm/Kmol x (23 + 273) K/ 0.96 atm = 0.038 L

(e ) When handling acids such as HCl it is required the use of safety goggles, acid resistant gloves and lab coat. It is also required to work under a safety hood since the vapors of HCl are toxic when inhaled.

To prepare 50.0 mL 2.0 M solution from the 12.3 M we will dilute it according to the following calculation:

V₁M₁ = V₂M₂ ⇒ V₁ = V₂M₂ /M₁

where V₁ is the volume of the 12.3 M HCl solution we are going to dilute, and V₂ is the 50.0 mL solution 2.0 M needed.

V₁ = 50.0 mL x 2.0 M / 12.3 M = 8.13 mL

Notice that in the above equation we do not need to convert the mL to L since V appears in both sides of the equation and will give us the volume in mL.

Now 8.13 mL is difficult to measure with a 10 ml graduated cylinder where we can read to 0.2 mL unless we accept the error.

So we need to calculate the mass of concentrated acid required by computing its density

We can calculate the density of the 12.3 M solution using a tared 10 mL graduated by taking say 10 mL of the the solution, weighting it, and calculating the density = mass of solution / volume.

Knowing the density we can calculate the mass of 12.3 M a volume of 8.13 mL weighs.

Place approximately 35 mL of distilled water in the volumetric flask and tare in the balance.

Add say 7 mL of 12.3 M HCl in the graduated cylinder to the volumetric flask being careful towards the end to add the last portions using the dropper to complete the required mass using the balance.

Finally dilute to the 50 mL mark.

Again use all of the safety precautions indicated above and avoid any contact of the acid with the skin.