Question

calculate the pressure in both atm and mmHg exerted if 20.0 mL of 0.10 M HCl reacts with excess magnesium in a 125 mL flask at 298 K. What mass of magnesium reacts? What would be the pressure if 1.0 M HCl is used? Assume the magnesium occupies no volume.

Answer 1

Answer –

Given, molarity of HCl = 0.10 M, volume = 20.0 mL , volume of flask = 125 mL, T = 298 K

Reaction –

2 HCl + Mg -------> MgCl₂ + H₂

First we need to calculate the moles of HCl

From the given molarity and volume

Moles of HCl = molarity * volume (L)

                       = 0.10 M * 0.020 L

                       = 0.002 moles of HCl

Now we are given that Mg is excess, so HCl is limiting reactant

So, moles of H₂ gas from the balanced equation

2 moles of HCl = 1 moles of H₂

0.002 moles of HCl = ?

= 0.002 moles of HCl * 1 moles of H₂ / 2 moles of HCl

= 0.001 moles of H₂

Now we have moles of H₂ gas, volume and temp, so we can calculate pressure using the Ideal gas law

PV= nRT

P = nRT/V

= 0.001 moles * 0.0821 L.atm.mol^-1.K^-1 * 298 K / 0.125 L

= 0.196 atm

Now we need to convert this pressure atm to mm Hg

We know

1 atm = 760 mm Hg

So, 0.196 atm = ?

= 148.7 mm Hg

Now when we use the 1.0 M HCl –

Moles of HCl = molarity * volume (L)

                       = 1.0 M * 0.020 L

                       = 0.02 moles of HCl

Now we are given that Mg is excess, so HCl is limiting reactant

So, moles of H₂ gas from the balanced equation

2 moles of HCl = 1 moles of H₂

0.02 moles of HCl = ?

= 0.02 moles of HCl * 1 moles of H₂ / 2 moles of HCl

= 0.01 moles of H₂

Using the Ideal gas law

PV= nRT

P = nRT/V

= 0.01 moles * 0.0821 L.atm.mol^-1.K^-1 * 298 K / 0.125 L

= 1.96 atm

Now we need to convert this pressure atm to mm Hg

We know

1 atm = 760 mm Hg

So, 1.96 atm = ?

= 1487 mm Hg