Question

A student used 0.0938 g of Mg and collected 96.21 mL of H2 gas (their eudiometer tube is larger than the
one we used) over water at 29.3°C on a day when the atmospheric pressure was 786.8 mmHg. The level
of the solution in the tube was 12.6 mm above the level in the beaker when the reaction was complete.
What was the experimental value for R in units of L mmHg mol1- K-1? How does this compare (calculate
percent error) to the accepted value for R?

Answer 1

Write the balanced chemical equation for the reaction.

Mg (s) + 2 HCl (aq) -------> MgCl₂ (aq) + H₂ (g)

As per the stoichiometric equation,

1 mole Mg = 1 mole H₂.

Atomic mass of Mg = 24.305 g/mol; molar mass of H₂ = 2*1.008 g/mol = 2.016 g/mol.

Mole(s) of Mg corresponding to 0.0938 g Mg = (0.0938 g)/(24.305 g/mol) = 0.0038593 mole.

Mole(s) of H₂ = (0.0038593 mole Mg)*(1 mole H₂/1 mole Mg) = 0.0038593 mole.

Use the ideal gas equation to find out the value of the gas constant, R.

P*V = n*R*T where P = 786.8 mmHg; V = 96.21 mL = (96.21 mL)*(1 L/1000 mL) = 0.09621 L; T = 29.3°C = (29.3 + 273.15) K = 302.45 K and n = 0.0038593 mole.

(786.8 mmHg)*(0.09621 L) = (0.0038593 mole)*R*(302.45 K)

====> R = 64.85186 L.mmHg/mol.K (ans).

The accepted value of R in L.mmHg/mol.K is 62.36367 L.mmHg/mol.K.

Percent error = │(accepted value) – (experimental value)│/(accepted value)*100 [│…│ denotes the absolute value without any sign]

= │(62.36367 L.mmHg/mol.K) – (64.85186 L.mmHg/mol.K)│/(62.36367 L.mmHg/mol.K)*100

= 3.9898% ≈ 4.00% (ans).