Answer:
(a) 0.0015 mol Mg
(b) 0.0030 mol HCl
(c) 728 torr
(d) 0.038 L
(e) See below
Explanation:
This problem is a calculation based on the stoichiometry for the reaction:
2 H⁺ (aq) + 2 Cl⁻ + Mg ⇒ Mg²⁺ (aq) + 2 Cl⁻ (aq) + H₂ (g)
Given the mass of Mg reacted, we have:
Atomic Weight Mg = 24.3 g/mol
(a) Mole Mg reacted = mass/AW = 0.0360 g/ 24.3 g/mol = 0.0015 mol
(b) Moles HCl needed:
2 mol HCl/ 1 mol Mg x 0.0015 mol Mg = 0.0030 mol HCl
(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.
