Question

The reaction is: Mg(s) + 2HCl(aq) --> MgCl2 (aq) + H2(g)

1. If 2.016 g of hydrogen (H2) was released, how many moles of magnesium reacted? How many grams of magnesium would this be?

2. How many moles of magnesium would replace one mole of hydrogen (H) in the hydrochloric acid? How many grams? (This is the equivalent weight)

3. Compare the mass calculated above in question 2 to your experimental equivalent weight and calculate the percent difference. What would cause your experimental value to be different? %difference = [(calculated value-experimental value)/(calculated value)]x100

4. Using the ideal gas law, calculate the value of R in L-atm/mol-K by assuming that an ideal gas occupies 22.4 L/mol at STP.

Answer 1

Mg + 2HCl ---------> MgCl₂ + H₂

1) Mass of hydrogen released = 2.016 g

Molar mass of hydrogen = 2 x 1.00794 g/mol = 2.01588 g/mol

Moles of hydrogen = Mass of hydrogen/molar mass of hydrogen = 2.016 g/2.01588 g/mol = 1.000059 mol

As Molar ratio of Magnesium and Hydrogen = 1:1 in the above equation

So no. of moles of magnesium required to produce 1 mole of hydrogen gas = 1mole

Molar mass of Magnesium = 24.305 g/mol

Mass of magnesium = moles of magnesium x molar mass of magnesium = 1.000059 mol x 24.305 g/mol = 24.306 gms of Magnesium

0.5 moles of Magnesium would be 12.153 g/equivalent weight

2) Molar ratio of Magnesium and Hydrogen chloride = 1:2 in the above equation

   0.5 Mg + 1 (H)Cl ----> 1/2 Mg⁺² + 0.5H₂ + 1 Cl^-

This shows that in order to replace 1 mole of hydrogen in hydrogen chloride (HCl) acid 0.5 moles (1/2) of Magnesium is required.

mass of magnesium - 24.305 g/mol

per 1 mole --------> 24.305 g

per 0.5 mole -------> ? 24.3 g x 0.5 mole/1 mole = 12.1525 gms-------/equivalent weight

3) Calculated/experimental value = 12.1525 gms

Theoretical value = 12.153 gms

Percent difference = [(experimental value-Theoretical value)/Theoretical value)] X 100 = [(12.1525-12.153 )/12.153] x 100 = 0.00411%

The difference would be that magnesium amount is in excess.

4) Ideal gas law PV = nRT

1 mole ------->22.4 lts

STP = Temperature = 273.15 K and pressure, P = 1 atm

1 atm x 22.4 lts = 1 mole x R X 273.15 K

22.4 lt atm/273.15 K mol = R

Ideal gas constant, R = 0.082006 lt atm/mol K